# Important Questions Class 12 Maths Chapter 7

## Important Questions for CBSE Class 12 Maths Chapter 7 – Integrals

Integrals are a generalisation of the area under a function’s curve in mathematical terms. They are included in the calculus topic, along with derivatives. It can also be represented numerically.

Class 12 Maths Chapter 7 Integration Important Questions were created by Extramarks subject matter experts to help students prepare for exams. This chapter teaches students about integral calculus (definite and indefinite), its properties, and much more. This subject is extremely important for both the CBSE board exam and competitive examinations. The concepts of integrals are explained thoroughly and clearly in this chapter. These Important Questions are straightforward and can easily assist students to problem-solve.

You can also find CBSE Class 12 Maths Important Questions Chapter-by-Chapter Important Questions here:

 CBSE Class 12 Maths Important Questions Sr No Chapter No Chapter Name 1 Chapter 1 Relations and Functions 2 Chapter 2 Inverse Trigonometric Functions 3 Chapter 3 Matrices 4 Chapter 4 Determinants 5 Chapter 5 Continuity and Differentiability 6 Chapter 6 Application of Derivatives 7 Chapter 7 Integrals 8 Chapter 8 Application of Integrals 9 Chapter 9 Differential Equations 10 Chapter 10 Vector Algebra 11 Chapter 11 Three Dimensional Geometry 12 Chapter 12 Linear Programming 13 Chapter 13 Probability

## Study Important Questions Class 12 Maths Chapter -7 Integrals

Class 12 Maths Chapter 7 Integration Important Questions follows the new NCERT guidelines and provides the students with the most recent questions and solutions. Students can resolve their doubts about each chapter by practising these important chapter questions and thorough explanations provided by our subject matter experts. These questions will aid students in their exam preparation and  then practise important questions quickly. The Integration Important Questions Class 12 Maths goes through all the topics part by part so that it is easy to prepare for the board exams.

Integration as an Inverse Differentiation Method

Integration is the inverse of differentiation. Instead of distinguishing a function, we are given its derivative and asked to find its primitive function, also known as the original function. This strategy is referred to as integration or anti-differentiation.

In fact, by arbitrarily selecting C from the real number set, there are an infinite number of anti-derivatives of each of these functions. C is commonly referred to as an arbitrary constant in this context. In fact, the parameter C is obtained by varying the function by which one obtains different antiderivatives (or integrals) of the given function.

Example:

Q. ∫(ax2+bx+c)dx

Solution:

∫(ax2+bx+c)dx=∫ax2 dx + ∫bxdx + ∫cdx

=a∫x2dx + b∫xdx+c∫dx

=ax2/3 +bx2/2 +cx+C

So, ∫(ax2+bx+c)dx=ax2/3 +bx2/2 +cx+C

Methods of Integration

The previous paragraph discussed integrals of those functions that were easily obtained from derivatives of certain functions. It was based on an inspection, i.e., looking for a function F whose derivative was f, which led us to the integral of f. However, for many functions, this inspection-based form is not appropriate. As a result of reducing them to standard types, we develop additional techniques or methods for locating the integrals. Among them, the following methods are popular:

1. Substitutional integration
2. Making Use of Partial Fraction Integration
3. Parts Integration

Example:

Q. Compute the Integral: ∫ex/2dx

Solution:

Let us equate, u=x/2

Differentiating both sides,

⇒du=dx/2

⇒dx=2/du

On integrating both sides,

⇒∫ex / 2dx = ∫eu⋅2du

⇒∫ex / 2dx = 2∫eudu

⇒∫ex / 2dx = 2eu + C

On putting the value of u=x/2, we get,

Answer: ∫ex / 2dx = 2ex / 2 + C

Integrals of Some Particular Functions

A rational function is defined as P(x) Q. (x) as the product of two polynomials, where P(x) and Q(x) are polynomials of x and Q(x), respectively. If the P(x) degree is less than the Q(x) degree, the rational function is said to be valid; otherwise, it is said to be improper.

Integration by Partial Fractions

If the integrand (the expression after the integral sign) is an algebraic fraction and the integral cannot be evaluated using simple methods, the fraction must be expressed in partial fractions before integration can occur. The steps required to decompose an algebraic fraction into its partial fractions result from considering the reverse process (or subtraction). We divide fractions into partial fractions because:

• It makes doing such integrals much easier.
• It is used in the Laplace transformation.

Before a fractional function can be directly represented in partial fractions, the numerator must be at least one degree lower than the denominator.

For each linear factor (ax+b) in the denominator of a rational fraction, there is a partial fraction of the form dfracA(ax+b), where A is a constant. If a linear factor is repeated n times in the denominator, there will be n corresponding partial fractions from degree 1 to n.

Integration by Parts

If u and v are two distinct functions of the same x variable (say). Then, according to the product law of differentiation, we have

dfracddx(uv) = udfracdvdx + vdfracdudx

The product integral of two functions = (first function) times (second function integral) – Integral of ((first function differential coefficient) times (second function integral)).

Definite Integral

The definite integral has its own meaning. A definite integral is denoted by int abf(x) dx, where an is taken into account.

an is considered the integral’s lower limit, and b is known as the integral’s upper limit. The definite integral is either added as the limit of a sum or, if it has an antiderivative F in the interval [a, b], its value is the difference between the F values at the endpoints, i.e. F(b)-F(a) (a).

Example:

∫1−1sin5x⋅cos4xdx

Solution:

Let, f(x)=sin5x⋅cos4xdx

f(−x)=sin5(−x)⋅cos−4(−x)dx

=−sin5x⋅cos4x

=−f(x)

From the above, it can be deduced that f(x) is an odd function.

Therefore, f(x)=∫1−1sin5x⋅cos4xdx=0

Conclusion

Important Integration Questions consist of Integrations, definite and indefinite integrals, certain definite integral properties, the fundamental calculus theorem, and integration methods such as:

• Integration by parts
• Integration by substitution
• Integration using partial fractions

The derivative concept is the focus of Differential Calculus. The original motivation for the derivative was the problem of defining tangent lines for function graphs and calculating the slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions.

As we all know, Class 12 marks are used to determine college admission. As a result, Extramarks is focused on providing students with various sets of questions that may be asked in the Class 12 Board Exams. At Extramarks, students will find Class 12 Maths Chapter 7 Integrals Important Questions that can be accessed at their convenience.

Q1.

$\mathrm{Evaluate}\frac{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}{\sqrt{1\text{}+\text{}\mathrm{sin}\text{}2\mathrm{x}}}\text{}\mathrm{dx}.$

Opt.

$\begin{array}{l}\frac{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}{\sqrt{1\text{}+\text{}\mathrm{sin}\text{}2\mathrm{x}}}\text{}\mathrm{dx}\\ =\frac{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}{\sqrt{{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}^{2}}}\text{}\mathrm{dx}\end{array}$

Ans.

$\begin{array}{l}«\frac{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}{\sqrt{1\text{}+\text{}\mathrm{sin}\text{}2\mathrm{x}}}\text{}\mathrm{dx}\\ =«\frac{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}{\sqrt{{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}^{2}}}\text{}\mathrm{dx}\\ =«\frac{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}{\left(\mathrm{sin}\text{}\mathrm{x}\text{}+\text{}\mathrm{cos}\text{}\mathrm{x}\right)}\text{}\mathrm{dx}\\ =«\mathrm{dx}\\ =\text{}\mathrm{x}\text{}+\text{}\mathrm{c}\end{array}$

Q2.

$\mathrm{Evaluate}\text{}\frac{{\mathrm{sin}}^{1}\mathrm{x}}{{\mathrm{sin}}^{1}\sqrt{\mathrm{x}}+\text{}{\mathrm{cos}}^{1}\sqrt{\mathrm{x}}}\text{}\mathrm{dx}.$

Opt.

$\begin{array}{l}\mathrm{We}\text{}\mathrm{have},\\ \frac{{\mathrm{sin}}^{1}\mathrm{x}}{{\mathrm{sin}}^{1}\sqrt{\mathrm{x}}+\text{}{\mathrm{cos}}^{1}\sqrt{\mathrm{x}}}\text{}\mathrm{dx}\\ \frac{{\mathrm{sin}}^{1}\mathrm{x}}{\frac{\mathrm{}}{2}}\text{}\mathrm{dx}\left[\text{}{\mathrm{sin}}^{1}\text{}\sqrt{\mathrm{x}}\right]\end{array}$

$\begin{array}{l}\left[\text{Ans.}\right]\end{array}$

Q3.

$\mathrm{Evaluate}\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{}}\mathrm{log}\text{}\mathrm{tanx}\text{}\mathrm{dx}.\text{}$

Opt.

$\begin{array}{l}\underset{}{\overset{}{Ans.}}\end{array}$

$\begin{array}{l}\mathrm{Let}\text{}\mathrm{I}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\mathrm{log}\text{}\mathrm{tanx}\text{}\mathrm{dx}\text{}.\dots .\left(1\right)\\ \mathrm{I}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\mathrm{log}\text{}\mathrm{tan}\left(\frac{\mathrm{}}{2}\text{}’\mathrm{x}\right)\text{}\mathrm{dx}\\ \mathrm{I}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\mathrm{log}\text{}\mathrm{cotx}\text{}\mathrm{dx}\text{}.\dots \left(2\right)\\ \mathrm{On}\text{}\mathrm{adding}\text{}\left(1\right)\text{}\mathrm{and}\text{}\left(2\right)\\ 2\mathrm{I}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\left(\mathrm{log}\text{}\mathrm{tanx}\text{}+\text{}\mathrm{log}\text{}\mathrm{cot}\text{}\mathrm{x}\right)\text{}\mathrm{dx}\\ 2\mathrm{I}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\left(\mathrm{log}\text{}\mathrm{tanx}.\mathrm{cot}\text{}\mathrm{x}\right).\text{}\mathrm{dx}\\ 2\mathrm{I}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\left(\mathrm{log}1\right).\text{}\mathrm{dx}\text{}=\text{}\underset{0}{\overset{\frac{\mathrm{}}{2}}{«}}\left(0\right).\text{}\mathrm{dx}\\ 2\mathrm{I}\text{}=\text{}0\\ \mathrm{I}\text{}=\text{}0\end{array}$