NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2

The subject of Mathematics is one of the most important in students’ academic careers. For students to succeed in the subject, they must practice meticulously. As Mathematics is a conceptual subject, students should also have a clear understanding of the subject in order to do well in their Class 12 board exams. Chapter 7 of Class 12 Mathematics is Integrals, and it is a topic that students might find challenging. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are a great resource for the students to understand the concepts of Chapter 7 Integrals. The solutions provide students with a consistent learning experience. Consequently, Extramarks allows students to access the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they can achieve high scores in their board examinations.

Class 12 is an extremely significant academic session for students. The academic session not only prepares them for competitive exams but also provides them with the foundation they need to succeed academically. The scores of Class 12 indicate the academic competence of the students. In addition, the scores of this academic session are also evaluated during the admissions to colleges, seeking job opportunities and much more. Therefore, Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they can practice and improve their academic performance.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.2) Exercise 7.2

Mathematics is a subject that many students may find challenging. The first and foremost step that students must take for the preparation of their Class 12 Mathematics board examination is to thoroughly go through the NCERT curriculum. However, the NCERT textbooks do not contain the solutions to all the questions included in them. Therefore, Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to help them prepare well for their examinations. The NCERT solutions provided by the Extramarks website help students review each step of the problem and understand the logic behind it. Practising the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 helps students solve problems accurately and at a better pace. This way, they can perform well in their examinations.

Students can easily download the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 from the Extramarks website. Mathematics is a subject that requires an ample amount of practice so that students can improve their conceptual clarity. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are considered one of the best resources for the preparation of the board examinations. They help students solve problems faster and more efficiently. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are easily accessible on any device as they are available in PDF format through the Extramarks website.

Important Properties and Formulas to Remember

Students can subscribe to the Extramarks website for complete and credible study material. There are various important properties and formulas in Class 12 Chapter 7 Integrals. It is very important for the students to readily remember these properties and formulas in order to perform well in the board examinations. Some of them are-

The methods of Differentiation and Integration are inverses of each other. Students can refer to the Extramarks website for a better understanding of this property.
Two Indefinite Integrals with the same derivative result in the same family of curves, and so, they are equivalent. Students can review the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to understand how these properties are practically applied to the questions.

There are multiple other formulas and functions that should be well-known to the students to be able to score better in the examinations. Students can refer to the Extramarks website to access the essential properties and formulas of Class 12 Chapter 7 Integrals. Extramarks is an organization that aims at the holistic development of students. It provides students with comprehensive study material so that students can succeed in any examination. NCERT content forms the conceptual academic base of students, therefore Extramarks offers students NCERT Solutions for all the subjects and classes. Students can refer to the Extramarks website for :

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NCERT Solutions for Class 12 Maths Chapter 7 – Exercise 7.2 Questions

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 provided by Extramarks are one of the best resources that help students prepare well to score high in board examinations. This is because the NCERT textbook covers all the topics that can appear in board examinations. Moreover, the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are curated by experienced and certified educationalists. NCERT textbooks are written by professionals who are experts in their subjects. Furthermore, they help students to understand the basic concepts of the subject, and they are ideal for students who find Mathematics difficult. Practising the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 helps students increase their pace of solving the problems of Mathematics, which is very essential for them to score well in the board examinations. Once students go through the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2, they will be able to solve any complicated problem that they may encounter in their examinations. Extramarks recommends students practice Class 12th Maths Chapter 7 Exercise 7.2 thoroughly prior to their examinations.

NCERT Solutions for Class 12 Maths Chapter 7 – Exercise 7.2 Questions

The major concepts involved in Class 12 Chapter 7 Integrals are the Introduction of the Chapter, Integration as an Inverse Process of Differentiation, Geometrical Interpretation of Indefinite Integral, Some Properties of Indefinite Integral, Comparison between Differentiation and Integration, and Methods of Integration. Other included topics are Integrals of Some Particular Functions, Integration by Partial Fractions, Integration by Partial Fractions, Integration by Parts, Definite Integral, Fundamental Theorem of Calculus, Evaluation of Definite Integrals by Substitution, and Some Properties of Definite Integrals. Exercise 7.2 Class 12 Maths Solutions are based on Integration by Substitution, Integration using Partial Fractions and Integration by Parts. Students can practice the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to structure their answers to Exercise 7.2 Class 12th.

Ex 7.2 Class 12 Maths NCERT Solutions: Introduction

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are based on the various Methods of Integration. Students can also practice the exemplar questions given before Exercise 7.2 to have a better understanding of the concepts of the exercise. Overall, there are eleven exercises in the chapter; therefore, it could be a huge challenge for the students to understand such a wide range of concepts. Therefore, Extramarks provides students with NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they can rigorously practice the NCERT textbook to be able to perform well in their examinations.

NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2: Formalization

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Exercise 7.2 Class 12 Maths NCERT Solutions

One of the best ways to improve the mathematical skills of students is to thoroughly review the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2. These solutions help students build strong fundamentals of the curriculum of the subject and also assist them in practising the application of those skills in real-life scenarios. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are designed for students to improve their mathematical skills quickly and effectively. Students can find the solutions complicated, but there are helpful resources to practice Chapter 7 Integrals. By revising with the help of the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 students can practice the concepts and functions of the chapter topics, which are very essential to scoring well in any in-school, board, or competitive examinations. Practising the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 is the primary step that students should take in preparation for their board examinations.

Class 12 Maths Ch Ex 7.2

In Mathematics, an integral can be defined as a numerical identity or a function of which the given function is a derivative. Furthermore, there are two types of Integrals – Definite and In-Definite Integrals. For a better understanding of the Chapter Integrals, students can refer to the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 and subscribe to the Extramarks website.

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Q1.

\int_{0}^{\frac{π}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \frac{sinx - cosx}{1 + sinxcosx} dx . .. (i) \\ = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - x) - \cos (\frac{π}{2} - x)}{1 + \sin (\frac{π}{2} - x) \cos (\frac{π}{2} - x)} dx \\ = \int_{0}^{\frac{π}{2}} \frac{cosx - sinx}{1 + cosxsinx} dx . .. (ii) \\ Adding equation (i) and equation (ii), we get \\ 2I = \int_{0}^{\frac{π}{2}} \frac{sinx - cosx + cosx - sinx}{1 + sinxcosx} dx \\ = \int_{0}^{\frac{π}{2}} \frac{0}{1 + sinxcosx} dx \\ = 0 \\ I = 0 \\ ∴ \int_{0}^{\frac{π}{2}} \frac{sinx - cosx}{1 + sinxcosx} dx = 0 \end{array}

Q2. ∫

0_{2 π} \cos^{5} x dx

Ans.

\begin{array}{l} Let I = \int_{0}^{2 π} \cos^{5} x dx \\ since, {cos}^{5} (2 π - x) = \cos^{5} x \\ ∵ \int_{0}^{2 a} f (x) dx = {\begin{cases} 2 \int_{0}^{a} f (x) dx, if f (2 a - x) = f (x) \\ 0, if f (2 a - x) = - f (x) \end{cases} \\ ∴ \int_{0}^{2 π} \cos^{5} x dx = 2 \int_{0}^{π} \cos^{5} x dx \\ = 0 [∵ \cos^{5} (π - x) = - \cos^{5} x] \end{array}

Q3.

\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{7} x dx

Ans.

\begin{array}{l} Let I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{7} x dx \\ Here, f (x) = \sin^{7} x and​ f (- x) = \sin^{7} (- x) \\ = - \sin^{7} x = - f (x) \\ So, ​ f (x) ​ is​​ an odd function, then \int_{- a}^{a} f (x) dx = 0 \\ ∴ I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{7} x dx = 0 \end{array}

Q4.

\int_{0}^{π} \frac{x}{1 + sinx} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{π} \frac{x}{1 + sinx} dx \\ = \int_{0}^{π} \frac{π - x}{1 + \sin (π - x)} dx \\ = \int_{0}^{π} \frac{π - x}{1 + sinx} dx \\ I = \int_{0}^{π} \frac{π}{1 + sinx} dx - \int_{0}^{π} \frac{x}{1 + sinx} dx \\ I = \int_{0}^{π} \frac{π}{1 + sinx} dx - I \\ 2 I = \int_{0}^{π} \frac{π (1 - sinx)}{(1 + sinx) (1 - sinx)} dx \\ = \int_{0}^{π} \frac{π (1 - sinx)}{1 - \sin^{2} x} dx \\ = π \int_{0}^{π} \frac{(1 - sinx)}{\cos^{2} x} dx \\ = π \int_{0}^{π} (\sec^{2} x - secxtanx) dx \\ = π {[tanx - secx]}_{0}^{π} \\ = π {(tanπ - secπ) - (\tan 0 - \sec 0)} \\ = π {0 - (- 1) - 0 + 1} \\ = 2 π \\ Ι = π \end{array}

Q5.

\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} x dx

Ans.

\begin{array}{l} Let I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} x dx \\ Here, f (x) = \sin^{2} x \\ so, f (- x) = \sin^{2} (- x) \\ = \sin^{2} x \\ = f (x) \\ ∴ I = 2 \int_{0}^{\frac{π}{2}} \sin^{2} x dx [\int_{- a}^{a} f (x) dx = 2 \int_{0}^{a} f (x) dx, if f (- x) = f (x)] \\ I = 2 \int_{0}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} dx \\ = \int_{0}^{\frac{π}{2}} (1 - \cos 2 x) dx \\ = {[x - \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} \\ = [\frac{π}{2} - \frac{1}{2} \sin 2 (\frac{π}{2})] - [0 - \frac{1}{2} \sin 2 (0)] \\ = \frac{π}{2} - \frac{1}{2} sinπ \\ = \frac{π}{2} \end{array}

Q6.

\int_{0}^{\frac{π}{2}} (2 \log \sin x - \log \sin 2 x) dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} (2 logsinx - logsin 2 x) dx \\ = \int_{0}^{\frac{π}{2}} (2 logsinx - \log 2 sinxcosx) dx \\ = \int_{0}^{\frac{π}{2}} (2 logsinx - \log 2 - logsinx - logcosx) dx \\ I = \int_{0}^{\frac{π}{2}} (logsinx - \log 2 - logcosx) dx . .. (i) \\ = \int_{0}^{\frac{π}{2}} {logsin (\frac{π}{2} - x) - \log 2 - logcos (\frac{π}{2} - x)} dx \\ [P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ = \int_{0}^{\frac{π}{2}} {logcosx - \log 2 - logsinx} dx . .. (ii) \\ Adding​ equation (i) and equation (ii), ​ we get \\ 2I = \int_{0}^{\frac{π}{2}} (logsinx - \log 2 - logcosx + logcosx - \log 2 - logsinx) dx \\ 2I = \int_{0}^{\frac{π}{2}} {- 2 \log 2} dx \\ = - 2 \log 2 {[x]}_{0}^{\frac{π}{2}} \\ = - 2 \log 2 [\frac{π}{2} - 0] \\ I = - (\frac{π}{2}) \log 2 = (\frac{π}{2}) \log \frac{1}{2} \end{array}

Q7.

\int_{0}^{2} x \sqrt{2 - x} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{2} x \sqrt{2 - x} dx \\ I = \int_{0}^{2} (2 - x) \sqrt{2 - (2 - x)} dx [P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ I = \int_{0}^{2} (2 - x) \sqrt{x} dx \\ I = \int_{0}^{2} (2 x^{\frac{1}{2}} - x^{\frac{3}{2}}) dx \\ = {[2 . \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}]}_{0}^{2} \\ = [2 . \frac{2^{\frac{3}{2}}}{\frac{3}{2}} - \frac{2^{\frac{5}{2}}}{\frac{5}{2}}] - [2 . \frac{0^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{0^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] \\ = [\frac{2^{2}}{3} . 2^{\frac{3}{2}} - \frac{2 {.2}^{\frac{5}{2}}}{5}] \\ = \frac{2^{\frac{7}{2}}}{3} - \frac{2^{\frac{7}{2}}}{5} \\ = 2^{\frac{7}{2}} (\frac{1}{3} - \frac{1}{5}) \\ = 2^{\frac{7}{2}} (\frac{5 - 3}{15}) \\ = 2^{\frac{7}{2}} (\frac{2}{15}) \\ = \frac{2^{\frac{9}{2}}}{15} \\ I = \frac{16 \sqrt{2}}{15} \end{array}

Q8.

\int_{0}^{\frac{π}{4}} \log (1 + tanx) dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{4}} \log (1 + tanx) dx \\ = \int_{0}^{\frac{π}{4}} \log {1 + \tan (\frac{π}{4} - x)} dx [P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ = \int_{0}^{\frac{π}{4}} \log {1 + \frac{\tan \frac{π}{4} - tanx}{1 + \tan \frac{π}{4} tanx}} dx \\ = \int_{0}^{\frac{π}{4}} \log {1 + \frac{1 - tanx}{1 + 1 . tanx}} dx \\ = \int_{0}^{\frac{π}{4}} \log {\frac{1 + tanx + 1 - tanx}{1 + tanx}} dx \\ = \int_{0}^{\frac{π}{4}} \log (\frac{2}{1 + tanx}) dx \\ I = \int_{0}^{\frac{π}{4}} \log 2 dx - \int_{0}^{\frac{π}{4}} \log (1 + tanx) dx \\ I = \log 2 {[x]}_{0}^{\frac{π}{4}} - I \\ 2 I = [\frac{π}{4} - 0] \log 2 \\ I = \frac{π}{8} \log 2 \end{array}

Q9.

\int_{0}^{1} x {(1 - x)}^{n} dx

Ans.

\begin{array}{l} Let  I = \int_{0}^{1} x {(1 - x)}^{n} dx \\ = \int_{0}^{1} (1 - x) {1 - (1 - x)}^{n} dx [P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ = \int_{0}^{1} (1 - x) x^{n} dx \\ = \int_{0}^{1} (x^{n} - x^{n + 1}) dx \\ = {[\frac{x^{n + 1}}{n + 1} - \frac{x^{n + 2}}{n + 2}]}_{0}^{1} \\ = [\frac{1^{n + 1}}{n + 1} - \frac{1^{n + 2}}{n + 2}] - [\frac{0^{n + 1}}{n + 1} - \frac{0^{n + 2}}{n + 2}] \\ = \frac{n + 2 - n - 1}{(n + 1) (n + 2)} = \frac{1}{(n + 1) (n + 2)} \end{array}

Q10.

\int_{2}^{8} | x - 5 | dx

Ans.

\begin{array}{l} Let  I = \int_{2}^{8} | x - 5 | dx \\ Since, ​ (x - 5) \leq 0 on [2, 5] and (x - 5) \geq 0 on [5, 8] . \\ ∴ I = \int_{2}^{5} - (x - 5) dx + \int_{5}^{8} (x - 5) dx \\ = {[- \frac{x^{2}}{2} + 5 x]}_{2}^{5} + {[\frac{x^{2}}{2} - 5 x]}_{5}^{8} \\ = [- \frac{5^{2}}{2} + 5 (5) - {- \frac{2^{2}}{2} + 5 (2)}] + [\frac{8^{2}}{2} - 5 (8) - {\frac{5^{2}}{2} - 5 (5)}] \\ = [- \frac{25}{2} + 25 + \frac{4}{2} - 10] + [\frac{64}{2} - 40 - \frac{25}{2} + 25] \\ = [\frac{25}{2} - 8] + [- 8 + \frac{25}{2}] \\ = 25 - 16 \\ = 9 \end{array}

Q11.

\int_{- 5}^{5} | x + 2 | dx

Ans.

\begin{array}{l} Let      I = \int_{- 5}^{5} | x + 2 | dx \\ Since, ​ (x + 2) \leq 0 on [- 5, - 2] and (x + 2) \geq 0 on [- 2, 5] . \\ ∴ I = \int_{- 5}^{- 2} - (x + 2) dx + \int_{- 2}^{5} (x + 2) dx \\ = {[- \frac{x^{2}}{2} - 2 x]}_{- 5}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{5} \\ = - [\frac{{(- 2)}^{2}}{2} + 2 (- 2) - \frac{{(- 5)}^{2}}{2} - 2 (- 5)] + [\frac{5^{2}}{2} + 2 (5) - \frac{{(- 2)}^{2}}{2} - 2 (- 2)] \\ = - [2 - 4 - \frac{25}{2} + 10] + [\frac{25}{2} + 10 - 2 + 4] \\ = - 8 + \frac{25}{2} + \frac{25}{2} + 12 \\ = 29 \end{array}

Q12.

\int_{0}^{\frac{π}{2}} \frac{\cos^{5} x}{\sin^{5} x + \cos^{5} x} dx

Ans.

\begin{array}{l} Let \\ I = \int_{0}^{\frac{π}{2}} \frac{\cos^{5} x}{\sin^{5} x + \cos^{5} x} dx . .. (i) \\ = \int_{0}^{\frac{π}{2}} \frac{\cos^{5} (\frac{π}{2} - x)}{\sin^{5} (\frac{π}{2} - x) + \cos^{5} (\frac{π}{2} - x)} dx [∵ P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ I = \int_{0}^{\frac{π}{2}} \frac{\sin^{5} x}{\cos^{5} x + \sin^{5} x} dx . .. (ii) \\ Adding  equation (i) and equation (ii), we get \\ 2 I = \int_{0}^{\frac{π}{2}} \frac{\cos^{5} x}{\sin^{5} x + \cos^{5} x} dx + \int_{0}^{\frac{π}{2}} \frac{\sin^{5} x}{\cos^{5} x + \sin^{5} x} dx \\ = \int_{0}^{\frac{π}{2}} \frac{\cos^{5} x + \sin^{5} x}{\sin^{5} x + \cos^{5} x} dx \\ = \int_{0}^{\frac{π}{2}} 1 . dx \\ = {[x]}_{0}^{\frac{π}{2}} \\ = \frac{π}{2} - 0 \\ = \frac{π}{2} \\ I = \frac{π}{4} \end{array}

Q13.

\int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx . .. (i) \\ = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} (\frac{π}{2} - x)}{\sin^{\frac{3}{2}} (\frac{π}{2} - x) + \cos^{\frac{3}{2}} (\frac{π}{2} - x)} dx \\ [∵ P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ I = \int_{0}^{\frac{π}{2}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} dx . .. (ii) \\ Adding  equation (i) and equation (ii), we get \\ 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx + \int_{0}^{\frac{π}{2}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} dx \\ = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} dx \\ = \int_{0}^{\frac{π}{2}} 1 . dx \\ = {[x]}_{0}^{\frac{π}{2}} \\ = \frac{π}{2} - 0 \\ = \frac{π}{2} \end{array}

Q14.

\int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{sinx}}{\sqrt{sinx} + \sqrt{cosx}} dx . .. (i) \\ = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} dx \\ [∵ P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx} + \sqrt{sinx}} dx . .. (ii) \\ Adding  equation (i) and equation (ii), we get \\ 2 I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{sinx}}{\sqrt{sinx} + \sqrt{cosx}} dx + \int_{0}^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx} + \sqrt{sinx}} dx \\ = \int_{0}^{\frac{π}{2}} \frac{\sqrt{sinx} + \sqrt{cosx}}{\sqrt{sinx} + \sqrt{cosx}} dx \\ = \int_{0}^{\frac{π}{2}} 1 . dx \\ = {[x]}_{0}^{\frac{π}{2}} \\ = \frac{π}{2} - 0 \\ = \frac{π}{2} \\ I = \frac{π}{4} \end{array}

Q15.

\int_{0}^{\frac{π}{2}} \cos^{2} x dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \cos^{2} x dx . .. (i) \\ = \int_{0}^{\frac{π}{2}} \cos^{2} (\frac{π}{2} - x) dx [∵ P_{4} : \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ I = \int_{0}^{\frac{π}{2}} {sin}^{2} x dx . .. (ii) \\ Adding equation (i) ​ and (ii),​ we get \\ 2I = \int_{0}^{\frac{π}{2}} (\cos^{2} x + {sin}^{2} x) dx \\ 2I = \int_{0}^{\frac{π}{2}} 1 dx [∵ \cos^{2} x + {sin}^{2} x = 1] \\ = {[x]}_{0}^{\frac{π}{2}} \\ = \frac{π}{2} - 0 \\ = \frac{π}{2} \\ I = \frac{π}{4} \end{array}

Q16.

\begin{array}{l} Choose the correct answer \\ If f (x) = \int_{0}^{x} t sint dt, then f ‘ (x) is \\ (A) cosx + x sinx \\ (B) x sinx \\ (C) x cosx \\ (D) sinx + x cosx \end{array}

Ans.

\begin{array}{l} We have f (x) = \int_{0}^{x} t sint dt \\ = {[t \int sint dt - \int \frac{d}{dt} t \int sint dt dt]}_{0}^{x} \\ = {[t (- cost) - \int 1 . (- cost) dt]}_{0}^{x} \\ = {[t (- cost) + \int cost dt]}_{0}^{x} \\ = {[t (- cost) + sint]}_{0}^{x} \\ = x (- cosx) + sinx - 0 (- \cos 0) - \sin 0 \\ = - xcosx + sinx \\ Differentiating​ w.r.t. x, we get \\ f’ (x) = - (x \frac{d}{dx} cosx + cosx \frac{d}{dx} x) + \frac{d}{dx} sinx \\ = - (x \times - sinx + cosx \times 1) + cosx \\ = xsinx - cosx + cosx \\ = xsinx \\ Hence, the correct option is B. \end{array}

Q17.

\begin{array}{l} Choose the correct answer \\ The value of the integral \int_{- \frac{1}{3}}^{1} \frac{{(x - x^{3})}^{\frac{1}{3}}}{x^{4}} dx is \\ (A) 6 \\ (B) 0 \\ (C) 3 \\ (D) 4 \end{array}

Ans.

\begin{array}{l} \int_{- \frac{1}{3}}^{1} \frac{{(x - x^{3})}^{\frac{1}{3}}}{x^{4}} dx \\ Let x = sinθ \Rightarrow \frac{dx}{dθ} = cosθ \\ When x = - \frac{1}{3}, θ = \sin^{- 1} (- \frac{1}{3}) and when x = 1, θ = \frac{π}{2} \\ \int_{- \frac{1}{3}}^{1} \frac{{(x - x^{3})}^{\frac{1}{3}}}{x^{4}} dx = \int_{\sin^{- 1} (- \frac{1}{3})}^{\frac{π}{2}} \frac{{(sinθ - \sin^{3} θ)}^{\frac{1}{3}}}{\sin^{4} θ} . cosθ dθ \\ = \int_{\sin^{- 1} (- \frac{1}{3})}^{\frac{π}{2}} \frac{{(sinθ)}^{\frac{1}{3}} {(1 - \sin^{2} θ)}^{\frac{1}{3}}}{\sin^{4} θ} . cosθ d θ \\ = \int_{\sin^{- 1} (- \frac{1}{3})}^{\frac{π}{2}} \frac{{(sinθ)}^{\frac{1}{3}} {(\cos^{2} θ)}^{\frac{1}{3}}}{\sin^{4} θ} . cosθ d θ \\ = \int_{\sin^{- 1} (- \frac{1}{3})}^{\frac{π}{2}} \frac{{(sinθ)}^{\frac{1}{3}} {(cosθ)}^{\frac{2}{3}}}{\sin^{2} θ . \sin^{2} θ} . cosθ dθ \\ = \int_{\sin^{- 1} (- \frac{1}{3})}^{\frac{π}{2}} \frac{{(cosθ)}^{\frac{5}{3}}}{\sin^{\frac{5}{3}} θ} . {cosec}^{2} θ dθ \\ = \int_{\sin^{- 1} (- \frac{1}{3})}^{\frac{π}{2}} {(cotθ)}^{\frac{5}{3}} . {cosec}^{2} θ dθ \\ Let t = cotθ \Rightarrow \frac{dt}{dθ} = - {cosec}^{2} θ \\ When θ = {sin}^{- 1} (- \frac{1}{3}), t = 2 \sqrt{2} and when θ = \frac{π}{2}, t = 0 \\ = - \int_{2 \sqrt{2}}^{0} {(t)}^{\frac{5}{3}} . dt = - {[\frac{t^{\frac{8}{3}}}{(\frac{8}{3})}]}_{2 \sqrt{2}}^{0} \\ = - \frac{3}{8} [0 - {(2 \sqrt{2})}^{\frac{8}{3}}] \\ = - \frac{3}{8} {(- 2^{\frac{3}{2}})}^{\frac{8}{3}} = \\ = \frac{3}{8} \times 2^{4} \\ = \frac{3}{8} \times 16 = 6 \\ Thus, the correct option is A. \end{array}

Q18.

Evaluate the integrals \int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x^{2}}) e^{2 x} dx

Ans.

\begin{array}{l} Let I = \int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x^{2}}) e^{2 x} dx \\ = \int_{1}^{2} (\frac{2}{2 x} - \frac{2}{{(2 x)}^{2}}) e^{2 x} dx \\ Again let ​ t = 2 x \Rightarrow \frac{dt}{dx} = 2 \\ When x = 1, t = 2 and when x = 2, t = 4 \\ I = \int_{2}^{4} (\frac{2}{t} - \frac{2}{t^{2}}) e^{t} \frac{dt}{2} \\ = \int_{2}^{4} (\frac{1}{t} - \frac{1}{t^{2}}) e^{t} dt \\ Let F (t) = \frac{1}{t}, Then F ‘ (t) = - \frac{1}{t^{2}} \\ \Rightarrow \int_{2}^{4} (\frac{1}{t} - \frac{1}{t^{2}}) e^{t} dt = \int_{2}^{4} (F (t) - F ‘ (t)) e^{t} dt \\ = {[e^{t} F (t)]}_{2}^{4} \\ = {[e^{t} . \frac{1}{t}]}_{2}^{4} \\ = e^{4} . \frac{2}{4} - e^{2} . \frac{1}{2} \\ = \frac{1}{4} e^{2} (e^{2} - 2) \\ Thus, \\ \int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x^{2}}) e^{2 x} dx = \frac{1}{4} e^{2} (e^{2} - 2) \end{array}

Q19.

Evaluate the integrals \int_{- 1}^{1} \frac{dx}{x^{2} + 2 x + 5} dx

Ans.

\begin{array}{l} \int_{- 1}^{1} \frac{dx}{x^{2} + 2 x + 5} dx = \int_{- 1}^{1} \frac{dx}{{(x + 1)}^{2} + 2^{2}} dx \\ Let t = x + 1 \Rightarrow \frac{dt}{dx} = 1 \\ When x = - 1, t = 0 ​  and when x = 1, t = 2 \\ ∴ \int_{- 1}^{1} \frac{dx}{{(x + 1)}^{2} + 2^{2}} dx = \int_{0}^{2} \frac{dx}{t^{2} + 2^{2}} dt \\ = {\frac{1}{2} \tan^{- 1} (\frac{t}{2})}_{0}^{2} \\ = \frac{1}{2} \tan^{- 1} (\frac{2}{2}) - \frac{1}{2} \tan^{- 1} (\frac{0}{2}) \\ = \frac{1}{2} \tan^{- 1} (1) - \frac{1}{2} \tan^{- 1} (0) \\ = \frac{1}{2} \times \frac{π}{4} - \frac{1}{2} \times 0 = \frac{π}{8} \\ Thus, \int_{- 1}^{1} \frac{dx}{x^{2} + 2 x + 5} dx = \frac{π}{8} \end{array}

Q20.

Evaluate the integrals \int_{0}^{2} \frac{dx}{x + 4 - x^{2}}

Ans.

\begin{array}{l} Let I = \int_{0}^{2} \frac{dx}{x + 4 - x^{2}} \\ = \int_{0}^{2} \frac{dx}{- (x^{2} - x - 4)} \\ = \int_{0}^{2} \frac{dx}{- (x^{2} - x + \frac{1}{4} - \frac{1}{4} - 4)} \\ = - \int_{0}^{2} \frac{dx}{{(x - \frac{1}{2})}^{2} - \frac{17}{4}} \\ = - \int_{0}^{2} \frac{dx}{{(x - \frac{1}{2})}^{2} - {(\frac{\sqrt{17}}{4})}^{2}} \\ Let t = x - \frac{1}{2} \Rightarrow \frac{dt}{dx} = 1 \\ When x = 0, t = - \frac{1}{2} and when x = 2, t = \frac{3}{2} \\ I = - \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{dt}{{(\frac{\sqrt{17}}{4})}^{2} - t^{2}} \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} \log {\frac{(\frac{\sqrt{17}}{4}) + t}{(\frac{\sqrt{17}}{4}) - t}}_{- \frac{1}{2}}^{\frac{3}{2}} \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} [\log {\frac{(\frac{\sqrt{17}}{4}) + \frac{3}{2}}{(\frac{\sqrt{17}}{4}) - \frac{3}{2}}} - \log {\frac{(\frac{\sqrt{17}}{4}) - \frac{1}{2}}{(\frac{\sqrt{17}}{4}) + \frac{1}{2}}}] \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} [\log {\frac{\sqrt{17} + 3}{\sqrt{17} - 3}} - \log {\frac{\sqrt{17} - 1}{\sqrt{17} + 1}}] \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} \log {\frac{\sqrt{17} + 3}{\sqrt{17} - 3} \times \frac{\sqrt{17} + 1}{\sqrt{17} - 1}} \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} \log {\frac{17 + \sqrt{17} + 3 \sqrt{17} + 3}{17 - \sqrt{17} - 3 \sqrt{17} + 3}} \\ = \frac{2}{\sqrt{17}} \log {\frac{20 + 4 \sqrt{17}}{20 - 4 \sqrt{17}}} \\ = \frac{2}{\sqrt{17}} \log {\frac{5 + \sqrt{17}}{5 - \sqrt{17}}} \\ = \frac{2}{\sqrt{17}} \log {\frac{5 + \sqrt{17}}{5 - \sqrt{17}} \times \frac{5 + \sqrt{17}}{5 + \sqrt{17}}} \\ = \frac{2}{\sqrt{17}} \log {\frac{25 + 17 + 10 \sqrt{17}}{25 - 17}} \\ = \frac{2}{\sqrt{17}} \log {\frac{42 + 10 \sqrt{17}}{8}} \\ = \frac{2}{\sqrt{17}} \log {\frac{21 + 5 \sqrt{17}}{4}} \end{array}

Q21.

Evaluate the integrals \int_{0}^{\frac{π}{2}} \frac{sinx}{1 + \cos^{2} x} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \frac{sinx}{1 + \cos^{2} x} dx \\ Again, let t = cosx \Rightarrow \frac{dt}{dx} = - sin x \\ When x = 0, t = 1 and  when x = \frac{π}{2}, t = 0 \\ ∴ I = \int_{1}^{0} \frac{sinx}{1 + t^{2}} \frac{dt}{- sinx} \\ = - \int_{1}^{0} \frac{1}{1 + t^{2}} dt \\ = - {[\tan^{- 1} t]}_{1}^{0} \\ = - (\tan^{- 1} 0 - \tan^{- 1} 1) \\ = - (0 - \frac{π}{4}) = \frac{π}{4} \end{array}

Q22.

Evaluate the integrals \int_{0}^{2} x \sqrt{x + 2} dx

Ans.

\begin{array}{l} I = \int_{0}^{2} x \sqrt{x + 2} dx \\ {Let​ t}^{2} = x + 2 \Rightarrow \frac{dt}{dx} = 2 t \\ When x = 0,  t = \sqrt{2} and when x = 2,  t = 2 \\ ∴ I = \int_{\sqrt{2}}^{2} (t^{2} - 2) \sqrt{t^{2}} 2 t dt \\ = 2 \int_{\sqrt{2}}^{2} (t^{4} - 2 t^{2}) dt \\ = 2 {[\frac{t^{5}}{5} - 2 \times \frac{t^{3}}{3}]}_{\sqrt{2}}^{2} \\ = 2 [\frac{2^{5}}{5} - 2 \times \frac{2^{3}}{3}] - 2 [\frac{{(\sqrt{2})}^{5}}{5} - 2 \times \frac{{(\sqrt{2})}^{3}}{3}] \\ = 2 [\frac{2^{5}}{5} - \frac{2^{4}}{3}] - 2 [\frac{4 \sqrt{2}}{5} - \frac{4 \sqrt{2}}{3}] \\ = 2^{5} [\frac{2}{5} - \frac{1}{3}] - 2^{3} \sqrt{2} [\frac{1}{5} - \frac{1}{3}] \\ = 2^{5} (\frac{6 - 5}{15}) - 2^{3} \sqrt{2} (\frac{3 - 5}{15}) \\ = \frac{2^{5}}{15} - 2^{3} \sqrt{2} (\frac{- 2}{15}) \\ = 2^{4} (\frac{2}{15} + \frac{\sqrt{2}}{15}) \\ = 16 \sqrt{2} (\frac{\sqrt{2} + 1}{15}) \end{array}

Q23.

Evaluate the integrals \int_{0}^{1} \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx

Ans.

\begin{array}{l} Let I = \int_{0}^{1} \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx \\ Also, let x = tan θ \Rightarrow \frac{dx}{dθ} = {sec}^{2} θ \\ When x = 0, θ = 0 and when x = 1, θ = \frac{π}{4} \\ ∴ I = \int_{0}^{\frac{π}{4}} {sin}^{- 1} (\frac{2 tan θ}{1 + {tan}^{2} θ}) {sec}^{2} θ d θ \\ = \int_{0}^{\frac{π}{4}} {sin}^{- 1} (sin 2 θ) {sec}^{2} θdθ \\ = \int_{0}^{\frac{π}{4}} 2 θ {sec}^{2} θ dθ \\ Now, \\ \int 2 θ {sec}^{2} θ dθ \\ = 2 [θ \int {sec}^{2} θ dθ - \int (\frac{d}{dθ} θ \int \sec^{2} θ dθ) dθ] \\ = 2 [θ tan θ - \int (1 . tan θ) dθ] \\ = 2 [θ tan θ + log cos θ] \\ ∴ I = 2 {[θ tan θ + log cos θ]}_{0}^{\frac{π}{4}} \\ = 2 [(\frac{π}{4}) tan \frac{π}{4} + log cos \frac{π}{4}] - [2 (0) tan 0 + log cos 0] \\ = 2 [\frac{π}{4} \times 1 + log \frac{1}{\sqrt{2}}] - [0 + log 1] \\ = [\frac{π}{2} - 2 \times \frac{1}{2} log 2] - [0 + 0] \\ = (\frac{π}{2} - log 2) \end{array}

Q24.

Evaluate the integrals \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} \cos^{5} ϕ dϕ

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} \cos^{5} ϕ d ϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} \cos^{4} ϕ cosϕ dϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} {(\cos^{2} ϕ)}^{2} cosϕ dϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} {(1 - \sin^{2} ϕ)}^{2} cosϕ dϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} (1 - 2 \sin^{2} ϕ + \sin^{4} ϕ) cosϕ dϕ \\ Let t = sinϕ \Rightarrow \frac{dt}{dϕ} = cosϕ \\ When ϕ = 0, t = 0 and when ϕ = \frac{π}{2}, t = 1 \\ ∴ I = \int_{0}^{1} \sqrt{t} (1 - 2 t^{2} + t^{4}) dt \\ = \int_{0}^{1} (t^{\frac{1}{2}} - 2 t^{\frac{5}{2}} + t^{\frac{9}{2}}) dt \\ = {[\frac{t^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{t^{\frac{7}{2}}}{\frac{7}{2}} + \frac{t^{\frac{11}{2}}}{\frac{11}{2}}]}_{0}^{1} \\ = [\frac{1^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{1^{\frac{7}{2}}}{\frac{7}{2}} + \frac{1^{\frac{11}{2}}}{\frac{11}{2}}] - [\frac{0^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{0^{\frac{7}{2}}}{\frac{7}{2}} + \frac{0^{\frac{11}{2}}}{\frac{11}{2}}] \\ = [\frac{2}{3} - 2 . \frac{2}{7} + \frac{2}{11}] \\ = \frac{2}{3} - \frac{4}{7} + \frac{2}{11} \\ = \frac{154 - 132 + 42}{231} = \frac{64}{231} \end{array}

Q25.

Evaluate the integrals \int_{0}^{1} \frac{x}{x^{2} + 1} dx

Ans.

\begin{array}{l} \int_{0}^{1} \frac{x}{x^{2} + 1} dx \\ Let t = x^{2} + 1 \Rightarrow \frac{dt}{dx} = 2 x \\ When x = 0, t = 1 and when x = 1,  t = 2 \\ ∴ \int_{0}^{1} \frac{x}{x^{2} + 1} dx = \frac{1}{2} \int_{1}^{2} \frac{1}{t} dt \\ = \frac{1}{2} {[\log | t |]}_{1}^{2} \\ = \frac{1}{2} [\log 2 - \log 1] \\ = \frac{1}{2} \log 2 \end{array}

Q26.

\begin{array}{l} Choose the correct answer \\ \int_{0}^{\frac{2}{3}} \frac{dx}{4 + 9 x^{2}} equals \\ (A) \frac{π}{6} \\ (B) \frac{π}{12} \\ (C) \frac{π}{24} \\ (D) \frac{π}{4} \end{array}

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{2}{3}} \frac{dx}{4 + 9 x^{2}} \\ \int \frac{1}{4 + 9 x^{2}} dx = \int \frac{1}{2^{2} + {(3 x)}^{2}} dx \\ = \frac{1}{2} \tan^{- 1} (\frac{3 x}{2}) \times \frac{1}{3} \\ = \frac{1}{6} \tan^{- 1} (\frac{3 x}{2}) = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{2}{3}) - F (0) \\ = \frac{1}{6} \tan^{- 1} (\frac{3}{2} \times \frac{2}{3}) - \frac{1}{6} \tan^{- 1} (\frac{3}{2} \times 0) \\ = \frac{1}{6} \tan^{- 1} (1) - \frac{1}{6} \tan^{- 1} (0) \\ = \frac{1}{6} \times \frac{π}{4} = \frac{π}{24} \\ Hence, the correct option is C. \end{array}

Q27.

\begin{array}{l} Choose the correct answer \\ \int_{1}^{\sqrt{3}} \frac{dx}{1 + x^{2}} equals \\ (A) \frac{π}{3} \\ (B) \frac{2 π}{3} \\ (C) \frac{π}{6} \\ (D) \frac{π}{12} \end{array}

Ans.

\begin{array}{l} Let I = \int_{1}^{\sqrt{3}} \frac{dx}{1 + x^{2}} \\ \int \frac{1}{1 + x^{2}} dx = \tan^{- 1} x \\ Therefore, by the second fundamental theorem, we have \\ I = F (\sqrt{3}) - F (1) \\ = {tan}^{- 1} (\sqrt{3}) - {tan}^{- 1} (1) \\ = \frac{π}{3} - \frac{π}{4} \\ = \frac{4 π - 3 π}{12} \\ = \frac{π}{12} \\ Hence, the correct option is D. \end{array}

Q28.

\int_{0}^{1} (x e^{x} + \sin \frac{πx}{4}) dx

Ans.

\begin{array}{l} Let I = \int_{0}^{1} (x e^{x} + \sin \frac{πx}{4}) dx \\ \int (x e^{x} + \sin \frac{πx}{4}) dx = \int x e^{x} dx + \int \sin \frac{πx}{4} dx \\ = x \int e^{x} dx - \int (\frac{d}{dx} x \int e^{x} dx) dx - \frac{1}{(\frac{π}{4})} \cos \frac{πx}{4} + C \\ = {xe}^{x} - \int e^{x} dx - \frac{4}{π} \cos \frac{πx}{4} \\ = {xe}^{x} - e^{x} - \frac{4}{π} \cos \frac{πx}{4} \\ = e^{x} (x - 1) - \frac{4}{π} \cos \frac{πx}{4} = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (0) \\ = {e^{1} (1 - 1) - \frac{4}{π} \cos \frac{π (1)}{4}} - {e^{0} (0 - 1) - \frac{4}{π} \cos \frac{π (0)}{4}} \\ = {- \frac{4}{π} \cos \frac{π}{4}} - {- 1 - \frac{4}{π} \times 1} \\ = {- \frac{4}{π} \times \frac{1}{\sqrt{2}}} - {- 1 - \frac{4}{π}} \\ = - \frac{4}{\sqrt{2} π} + 1 + \frac{4}{π} \\ = 1 + \frac{4}{π} - \frac{2 \sqrt{2}}{π} \end{array}

Q29.

\int_{0}^{2} \frac{6 x + 3}{x^{2} + 4} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{2} \frac{6 x + 3}{x^{2} + 4} dx \\ \int \frac{6 x + 3}{x^{2} + 4} dx = \int \frac{6 x}{x^{2} + 4} dx + \int \frac{3}{x^{2} + 4} dx \\ = 3 \int \frac{2 x}{x^{2} + 4} dx + 3 \int \frac{1}{x^{2} + 2^{2}} dx \\ = 3 [\log (x^{2} + 4)] + \frac{3}{2} [\tan^{- 1} \frac{x}{2}] = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (2) - F (0) \\ = 3 [\log (2^{2} + 4) - \frac{3}{2} \tan^{- 1} \frac{2}{2}] - [\log (0^{2} + 4) + \frac{3}{2} \tan^{- 1} \frac{0}{2}] \\ = 3 [\log 8 - \log 4] - \frac{3}{2} [\tan^{- 1} (1) - \tan^{- 1} (0)] \\ = 3 \log \frac{8}{4} - \frac{3}{2} [\frac{π}{4} - 0] \\ \int_{0}^{2} \frac{6 x + 3}{x^{2} + 4} dx = 3 \log 2 - \frac{3 π}{8} \end{array}

Q30.

\int_{0}^{π} (\sin^{2} \frac{x}{2} - \cos^{2} \frac{x}{2}) dx

Ans.

\begin{array}{l} Let I = \int_{0}^{π} (\sin^{2} \frac{x}{2} - \cos^{2} \frac{x}{2}) dx \\ \int (\sin^{2} \frac{x}{2} - \cos^{2} \frac{x}{2}) dx = - \int (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}) dx \\ = - \int cosx dx [∵ \cos^{2} θ - \sin^{2} θ = \cos 2 θ] \\ = - sinx = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (π) - F (0) \\ = - sinπ - (- \sin 0) = 0 \end{array}

Q31.

\int_{0}^{\frac{π}{4}} (2 \sec^{2} x + x^{3} + 2) dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{4}} (2 \sec^{2} x + x^{3} + 2) dx \\ \int (2 \sec^{2} x + x^{3} + 2) dx \\ = 2 \int \sec^{2} x dx + \int x^{3} dx + \int 2 dx \\ = 2 tanx + \frac{x^{4}}{4} + 2 x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{4}) - F (0) \\ = {2 \tan (\frac{π}{4}) + \frac{{(\frac{π}{4})}^{4}}{4} + 2 (\frac{π}{4})} - {2 \tan (0) + \frac{{(0)}^{4}}{4} + 2 (0)} \\ = 2 [\tan \frac{π}{4} - \tan 0] + \frac{1}{4} [{(\frac{π}{4})}^{4} - 0] + 2 [\frac{π}{4} - 0] \\ = 2 [1 - 0] + \frac{1}{4} [{(\frac{π}{4})}^{4}] + 2 [\frac{π}{4}] \\ = 2 + \frac{π^{4}}{1024} + \frac{π}{2} \end{array}

Q32.

\int_{1}^{2} \frac{5 x^{2}}{x^{2} + 4 x + 3} dx

Ans.

\begin{array}{l} Let                 I = \int_{1}^{2} \frac{5 x^{2}}{x^{2} + 4 x + 3} dx \\ \int \frac{5 x^{2}}{x^{2} + 4 x + 3} dx = \int (5 - \frac{20 x + 15}{x^{2} + 4 x + 3}) dx \\ = \int 5 dx - \int \frac{20 x + 15}{x^{2} + 4 x + 3} dx . .. (i) \\ Let 20 x + 15 = A \frac{d}{dx} (x^{2} + 4 x + 3) + B \\ = A (2 x + 4) + B \\ = 2 Ax + 4 A + B \\ Equating the coefficients of x and constant term, we get \\ \Rightarrow 2 A = 20 and 4A + B = 15 \\ \Rightarrow A = 10 and B = - 25 \\ ∴ 20 x + 15 = 10 (2 x + 4) - 25 \\ From equation (i), we have \\ \int \frac{5 x^{2}}{x^{2} + 4 x + 3} dx = \int 5 dx - \int \frac{10 (2 x + 4) - 25}{x^{2} + 4 x + 3} dx \\ = \int 5 dx - 10 \int \frac{(2 x + 4)}{x^{2} + 4 x + 3} dx + 25 \int \frac{1}{x^{2} + 4 x + 3} dx \\ = 5 x - 10 \log (x^{2} + 4 x + 3) + 25 \int \frac{1}{{(x + 2)}^{2} - 1} dx \\ = 5 x - 10 \log (x^{2} + 4 x + 3) + 25 \times \frac{1}{2} \log | \frac{x + 2 - 1}{x + 2 + 1} | + C \\ = 5 x - 10 \log (x^{2} + 4 x + 3) + \frac{25}{2} \log | \frac{x + 1}{x + 3} | \\ \int_{1}^{2} \frac{5 x^{2}}{x^{2} + 4 x + 3} dx = {[5 x - 10 \log (x^{2} + 4 x + 3) + \frac{25}{2} \log | \frac{x + 1}{x + 3} |]}_{1}^{2} \\ = [5 (2) - 10 \log (2^{2} + 4 \times 2 + 3) + \frac{25}{2} \log | \frac{2 + 1}{2 + 3} |] \\ - [5 (1) - 10 \log (1^{2} + 4 \times 1 + 3) + \frac{25}{2} \log | \frac{1 + 1}{1 + 3} |] \\ \int_{1}^{2} \frac{5 x^{2}}{x^{2} + 4 x + 3} dx = {[5 x - 10 \log (x^{2} + 4 x + 3) + \frac{25}{2} \log | \frac{x + 1}{x + 3} |]}_{1}^{2} \\ = [5 (2) - 10 \log (2^{2} + 4 \times 2 + 3) + \frac{25}{2} \log | \frac{2 + 1}{2 + 3} |] \\ - [5 (1) - 10 \log (1^{2} + 4 \times 1 + 3) + \frac{25}{2} \log | \frac{1 + 1}{1 + 3} |] \\ = (10 - 10 \log 15 + \frac{25}{2} \log \frac{3}{5}) - (5 - 10 \log 8 + \frac{25}{2} \log \frac{2}{4}) \\ = 5 - 10 (\log 15 - \log 8) + \frac{25}{2} (\log \frac{3}{5} - \log \frac{2}{4}) \\ = 5 - 10 (\log 5 + \log 3 - \log 4 - \log 2) \\ + \frac{25}{2} (\log 3 - \log 5 - \log 2 + \log 4) \\ = 5 - (10 + \frac{25}{2}) \log 5 + (10 + \frac{25}{2}) \log 4 + (- 10 + \frac{25}{2}) \log 3 \\ + (10 - \frac{25}{2}) \log 2 \\ = 5 - \frac{45}{2} \log 5 + \frac{45}{2} \log 4 + \frac{5}{2} \log 3 - \frac{5}{2} \log 2 \\ \int_{1}^{2} \frac{5 x^{2}}{x^{2} + 4 x + 3} dx = 5 - \frac{45}{2} \log \frac{5}{4} + \frac{5}{2} \log \frac{3}{2} = 5 - \frac{5}{2} (9 \log \frac{5}{4} - \log \frac{3}{2}) \end{array}

Q33.

\int_{0}^{1} x e^{x^{2}} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{1} x e^{x^{2}} dx \\ ∵ \int x e^{x^{2}} dx = \int x e^{t} \frac{dt}{2 x} [Let​ t = x^{2} \Rightarrow \frac{dt}{dx} = 2 x \Rightarrow dx = \frac{dt}{2 x}] \\ = \frac{1}{2} e^{t} \\ = \frac{1}{2} e^{x^{2}} = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (0) \\ = \frac{1}{2} e^{1^{2}} - \frac{1}{2} e^{0} = \frac{1}{2} (e - 1) \end{array}

Q34.

\int_{0}^{1} \frac{2 x + 3}{(5 x^{2} + 1)} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{1} \frac{2 x + 3}{(5 x^{2} + 1)} dx \\ = \frac{1}{5} \int_{0}^{1} \frac{10 x}{(5 x^{2} + 1)} dx + \int_{0}^{1} \frac{3}{(5 x^{2} + 1)} dx \\ \int \frac{2 x + 3}{(5 x^{2} + 1)} dx = \frac{1}{5} \int \frac{10 x}{(5 x^{2} + 1)} dx + 3 \int \frac{1}{(5 x^{2} + 1)} dx \\ = \frac{1}{5} \log | 5 x^{2} + 1 | + 3 \int \frac{1}{{(\sqrt{5} x)}^{2} + 1} dx \\ = \frac{1}{5} \log | 5 x^{2} + 1 | + 3 \tan^{- 1} \sqrt{5} x . \frac{1}{\sqrt{5}} \\ = \frac{1}{5} \log | 5 x^{2} + 1 | + \frac{3}{\sqrt{5}} \tan^{- 1} (\sqrt{5} x) = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (0) \\ = \frac{1}{5} \log | 5 {(1)}^{2} + 1 | + \frac{3}{\sqrt{5}} \tan^{- 1} {\sqrt{5} (1)} - [\begin{array}{l} \frac{1}{5} \log | 5 {(0)}^{2} + 1 | \\ + \frac{3}{\sqrt{5}} \tan^{- 1} {\sqrt{5} (0)} \end{array}] \\ = \frac{1}{5} \log (6) + \frac{3}{\sqrt{5}} \tan^{- 1} (\sqrt{5}) - \frac{1}{5} \log | 1 | - \frac{3}{\sqrt{5}} \tan^{- 1} (0) \\ = \frac{1}{5} \log (6) + \frac{3}{\sqrt{5}} \tan^{- 1} (\sqrt{5}) \end{array}

Q35.

\int_{2}^{3} \frac{x dx}{(x^{2} + 1)}

Ans.

\begin{array}{l} Let I = \int_{2}^{3} \frac{x dx}{(x^{2} + 1)} \\ ∵ \int \frac{x}{(x^{2} + 1)} dx = \frac{1}{2} \log | x^{2} + 1 | = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (3) - F (2) \\ = \frac{1}{2} \log | 3^{2} + 1 | - \frac{1}{2} \log | 2^{2} + 1 | \\ = \frac{1}{2} \log | 10 | - \frac{1}{2} \log | 5 | \\ = \frac{1}{2} \log | \frac{10}{5} | \\ = \frac{1}{2} \log | 2 | \end{array}

Q36.

\int_{0}^{\frac{π}{2}} \cos^{2} x dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \cos^{2} x dx = \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} dx \\ ∵ \int \frac{1 + \cos 2 x}{2} dx = \frac{1}{2} (x + \frac{\sin 2 x}{2}) = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{2}) - F (0) \\ = \frac{1}{2} {\frac{π}{2} + \frac{\sin 2 . (\frac{π}{2})}{2}} - \frac{1}{2} {0 + \frac{\sin 2 (0)}{2}} \\ = \frac{1}{2} (\frac{π}{2} + \frac{sinπ}{2}) - \frac{1}{2} {0 + 0} \\ = \frac{1}{2} (\frac{π}{2}) = \frac{π}{4} \end{array}

Q37.

\int_{2}^{3} \frac{dx}{(x^{2} - 1)}

Ans.

\begin{array}{l} Let I = \int_{2}^{3} \frac{dx}{(x^{2} - 1)} \\ ∵ \int \frac{1}{x^{2} - 1} dx = \frac{1}{2} \log | \frac{x - 1}{x + 1} | = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (3) - F (2) \\ = \frac{1}{2} \log | \frac{3 - 1}{3 + 1} | - \frac{1}{2} \log | \frac{2 - 1}{2 + 1} | \\ = \frac{1}{2} \log | \frac{2}{4} | - \frac{1}{2} \log | \frac{1}{3} | \\ = \frac{1}{2} \log | \frac{2}{4} \times \frac{3}{1} | \\ = \frac{1}{2} \log | \frac{3}{2} | \end{array}

Q38.

\int_{0}^{1} \frac{1}{(1 + x^{2})} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{1} \frac{1}{1 + x^{2}} dx \\ ∵ \int \frac{1}{1 + x^{2}} dx = \tan^{- 1} x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (0) \\ = \tan^{- 1} (1) - \tan^{- 1} (0) \\ = \frac{π}{4} - 0 = \frac{π}{4} \end{array}

Q39.

\int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} dx

Ans.

\begin{array}{l} Let I = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} dx \\ ∵ \int \frac{1}{\sqrt{1 - x^{2}}} dx = \sin^{- 1} x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (0) \\ = \sin^{- 1} (1) - \sin^{- 1} (0) \\ = \frac{π}{2} - 0 = \frac{π}{2} \end{array}

Q40.

\int_{\frac{π}{6}}^{\frac{π}{4}} cosec x dx

Ans.

\begin{array}{l} Let I = \int_{\frac{π}{6}}^{\frac{π}{4}} cosec x dx \\ ∵ \int cosec x dx = \log | cosecx - cotx | = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{4}) - F (\frac{π}{6}) \\ = \log | cosec \frac{π}{4} - \cot \frac{π}{4} | - \log | cosec \frac{π}{6} - \cot \frac{π}{6} | \\ = \log | \sqrt{2} - 1 | - \log | 2 - \sqrt{3} | \\ = \log (\frac{\sqrt{2} - 1}{2 - \sqrt{3}}) \end{array}

Q41.

\int_{0}^{\frac{π}{4}} \tan x dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{4}} \tan x dx \\ ∵ \int \tan x dx = - \log | cosx | = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{4}) - F (0) \\ = - \log | \cos \frac{π}{4} | - (- \log | \cos 0 |) \\ = - \log | \cos \frac{π}{4} | + \log | \cos 0 | \\ = - \log | \frac{1}{\sqrt{2}} | + \log 1 \\ = \log \sqrt{2} + 0 = \frac{1}{2} \log 2 \end{array}

Q42.

\int_{4}^{5} e^{x} dx

Ans.

\begin{array}{l} Let I = \int_{4}^{5} e^{x} dx \\ ∵ \int e^{x} dx = e^{x} = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (5) - F (4) \\ = e^{5} - e^{4} \\ = e^{4} (e - 1) \end{array}

Q43.

\int_{0}^{\frac{π}{2}} \cos 2 x dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \cos 2 x dx \\ ∵ \int \cos 2 x dx = \frac{1}{2} \sin 2 x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{2}) - F (0) \\ = - \frac{1}{2} \sin 2 (\frac{π}{2}) - (- \frac{1}{2} \sin 0) \\ = - \frac{1}{2} sinπ + \frac{1}{2} \sin 0 \\ = 0 + 0 = 0 \end{array}

Q44.

\int_{0}^{\frac{π}{4}} \sin 2 x dx

Ans.

\begin{array}{l} Let I = \int_{0}^{\frac{π}{4}} \sin 2 x dx \\ ∵ \int \sin 2 x dx = - \frac{1}{2} \cos 2 x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{4}) - F (0) \\ = - \frac{1}{2} \cos 2 (\frac{π}{4}) - (- \frac{1}{2} \cos 0) \\ = - \frac{1}{2} \cos \frac{π}{2} + \frac{1}{2} \cos 0 \\ = 0 + \frac{1}{2} (1) = \frac{1}{2} \end{array}

Q45.

\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) dx

Ans.

\begin{array}{l} Let I = \int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) dx \\ ∵ \int (4 x^{3} - 5 x^{2} + 6 x + 9) dx \\ = \int 4 x^{3} dx - \int 5 x^{2} dx + \int 6 x dx + \int 9 dx \\ = x^{4} - \frac{5}{3} x^{3} + 3 x^{2} + 9 x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (2) - F (1) \\ = {2^{4} - \frac{5}{3} {(2)}^{3} + 3 {(2)}^{2} + 9 (2)} - {{(1)}^{4} - \frac{5}{3} {(1)}^{3} + 3 {(1)}^{2} + 9 (1)} \\ = (16 - \frac{40}{3} + 12 + 18) - (1 - \frac{5}{3} + 3 + 9) \\ = (46 - \frac{40}{3}) - (13 - \frac{5}{3}) \\ = 46 - \frac{40}{3} - 13 + \frac{5}{3} \\ = 33 - \frac{35}{3} \\ = \frac{64}{3} \end{array}

Q46.

\int_{2}^{3} \frac{1}{x} dx

Ans.

\begin{array}{l} Let I = \int_{2}^{3} \frac{1}{x} dx \\ ∵ \int \frac{1}{x} dx = logx = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (3) - F (2) \\ = \log 3 - \log 2 = \log \frac{3}{2} \end{array}

Q47.

\int_{- 1}^{1} (x + 1) dx

Ans.

\begin{array}{l} Let I = \int_{- 1}^{1} (x + 1) dx \\ ∵ \int (x + 1) dx = \frac{x^{2}}{2} + x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (- 1) \\ = (\frac{1^{2}}{2} + 1) - {\frac{{(- 1)}^{2}}{2} + (- 1)} \\ = \frac{3}{2} - (- \frac{1}{2}) \\ = 2 \end{array}

Q48.

Evaluate the following definite integrals as limit of sums . \int_{0}^{4} (x + e^{2 x}) dx

Ans.

\begin{array}{l} We have, \int_{0}^{4} (x + e^{2 x}) dx = \int_{0}^{4} x dx + \int_{0}^{4} e^{2 x} dx \\ = I_{1} + I_{2} (Let) \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)] \\ where, h = \frac{b - a}{n} \\ Here, a = 0, b = 4, f (x) = x, h = \frac{4 - 0}{n} = \frac{4}{n} \\ I_{1} = \int_{0}^{4} x dx \\ ∴ \int_{0}^{4} x dx = (4 - 0) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (0 + h) + . .. + f (0 + (n - 1) h)] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [0 + h + . .. + (n - 1) h] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [h + 2 h + . .. + (n - 1) h] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [{1 + 2 + . .. + (n - 1)} h] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = 4 \lim_{n \to \infty} [\frac{(n - 1)}{2} (\frac{4}{n})] \\ = 4 \lim_{n \to \infty} [\frac{4}{2} (1 - \frac{1}{n})] \\ = 4 [2 (1 - 0)] \\ = 4 (2) = 8 \\ I_{2} = \int_{0}^{4} e^{2 x} dx \\ = (4 - 0) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (0 + h) + . .. + f (0 + (n - 1) h)] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [e^{0} + e^{2 h} + . .. + e^{2 (n - 1) h}] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [1 + e^{2 h} + . .. + e^{2 (n - 1) h}] \\ Using the sum to n terms of a G.P., where a = 1, r = e^{2 h}, \\ we have \end{array}

\begin{array}{l} ∴ \int_{0}^{4} e^{2 x} dx = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{1 . (e^{2 nh} - 1)}{(e^{2 h} - 1)}] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(e^{2 n . \frac{4}{n}} - 1)}{(e^{\frac{8}{n}} - 1)}] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(e^{8} - 1)}{(e^{\frac{8}{n}} - 1)}] \\ = \frac{4 (e^{8} - 1)}{\lim_{n \to \infty} [\frac{e^{\frac{8}{n}} - 1}{\frac{8}{n}}] . 8} \\ = \frac{4 (e^{8} - 1)}{8} [∵ \lim_{x \to 0} \frac{e^{h} - 1}{h} = 1] \\ = \frac{e^{8} - 1}{2} \\ So, \int_{0}^{4} (x + e^{2 x}) dx \\ = I_{1} + I_{2} \\ = 8 + \frac{e^{8} - 1}{2} \\ = \frac{15 + e^{8}}{2} \end{array}

Q49.

Evaluate the following definite integrals as limit of sums . \int_{- 1}^{1} e^{x} dx

Ans.

\begin{array}{l} We have, \int_{- 1}^{1} e^{x} dx \\ ∵ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, a = - 1, b = 1, f (x) = e^{x}, h = \frac{1 - (- 1)}{n} = \frac{2}{n} \\ ∴ \int_{- 1}^{1} e^{x} dx = (1 + 1) \lim_{n \to \infty} \frac{1}{n} [f (- 1) + f (- 1 + h) + . .. + f (- 1 + (n - 1) h)] \\ = 2 \lim_{n \to \infty} \frac{1}{n} [e^{- 1} + e^{- 1 + h} + . .. + e^{- 1 + (n - 1) h}] \\ Using the sum to n terms of a G.P., where a = e^{- 1}, r = e^{h}, \\ we have \\ ∴ \int_{- 1}^{1} e^{x} dx = 2 \lim_{n \to \infty} \frac{1}{n} [\frac{e^{- 1} (e^{nh} - 1)}{(e^{h} - 1)}] \\ = 2 \lim_{n \to \infty} \frac{1}{n} [\frac{e^{- 1} (e^{n . \frac{2}{n}} - 1)}{(e^{\frac{2}{n}} - 1)}] \\ = 2 \lim_{n \to \infty} \frac{1}{n} [\frac{e^{- 1} (e^{2} - 1)}{(e^{\frac{2}{n}} - 1)}] \\ = \frac{2 e^{- 1} (e^{2} - 1)}{\lim_{n \to \infty} [\frac{e^{\frac{2}{n}} - 1}{\frac{2}{n}}] . 2} \\ = \frac{2 e^{- 1} (e^{2} - 1)}{2} \\ = e - \frac{1}{e} [∵ \lim_{x \to 0} \frac{e^{h} - 1}{h} = 1] \end{array}

Q50.

Evaluate the following definite integrals as limit of sums . \int_{1}^{4} (x^{2} - x) dx

Ans.

\begin{array}{l} We have, \int_{1}^{4} (x^{2} - x) dx = \int_{1}^{4} x^{2} dx - \int_{1}^{4} x dx \\ = I_{1} - I_{2} (Let) \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)] \\ where, h = \frac{b - a}{n} \\ Here, a = 1, b = 4, f (x) = x^{2}, h = \frac{4 - 1}{n} = \frac{3}{n} \\ ∴ I_{1} = \int_{1}^{4} x^{2} dx \\ = (4 - 1) \lim_{n \to \infty} \frac{1}{n} [f (1) + f (1 + h) + . .. + f (1 + (n - 1) h)] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [{(1)}^{2} + {(1 + h)}^{2} + . .. + {1 + (n - 1) h}^{2}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [1 + (1 + 2 h + h^{2}) + . .. + {1 + 2 (n - 1) h + {(n - 1)}^{2} h^{2}}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + {1 + 2 + . .. + (n - 1)} 2 h + {1 + 2^{2} + 3^{2} + . .. + {(n - 1)}^{2}} h^{2}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) n}{2} 2 h + \frac{(n - 1) n (2 n - 2 + 1)}{6} h^{2}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + (n - 1) n . \frac{3}{n} + \frac{(n - 1) n (2 n - 2 + 1)}{6} . \frac{9}{n^{2}}] \\ [∵ \sum n = \frac{n (n + 1)}{2}, \sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] \\ = 3 \lim_{n \to \infty} [1 + (1 - \frac{1}{n}) \times 3 + \frac{(1 - \frac{1}{n}) (2 - \frac{1}{n})}{6} \times 9] \end{array}

\begin{array}{l} = 3 [1 + 3 (1 - 0) + \frac{3 (1 - 0) (2 - 0)}{2}] \\ = 3 (1 + 3 + 3) \\ = 21 \\ For {​   I}_{2} = \int_{1}^{4} x dx \\ ∴ \int_{1}^{4} x dx = (4 - 1) \lim_{n \to \infty} \frac{1}{n} [f (1) + f (1 + h) + . .. + f (1 + (n - 1) h)] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [1 + 1 + h + . .. + 1 + (n - 1) h] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + h + 2 h + . .. + (n - 1) h] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + {1 + 2 + . .. + (n - 1)} h] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = 3 \lim_{n \to \infty} [1 + \frac{(n - 1)}{2} (\frac{3}{n})] \\ = 3 \lim_{n \to \infty} [1 + \frac{3}{2} (1 - \frac{1}{n})] \\ = 3 [1 + \frac{3}{2} (1 - 0)] \\ = 3 (\frac{5}{2}) = \frac{15}{2} \\ So, \int_{1}^{4} (x^{2} - x) dx \\ = I_{1} - I_{2} = 21 - \frac{15}{2} = \frac{27}{2} \end{array}

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NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2

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