What is the difference between definite and indefinite integrals?

When the upper and lower limits of an integral are specified, it is referred to as a definite integral. When calculating the definite integration of a given function f(x), keep in mind that you are calculating the area under the curve f(x) from x=a to x=b for upper and lower limits a and b. Indefinite integrals, unlike definite integrals, do not have upper and lower bounds. So you compute a generic integral value in indefinite integral for a family of similar functions f(x), where 'x' can have a variety of solutions. There is also a notation for the additive constant, C, which is written alongside the indefinite integral value of any function.

What do you mean by differentiation?

Differentiation is a method for calculating the derivative of a function. Differentiation is a mathematical procedure for determining a function's instantaneous rate of change based on one of its variables. The most obvious example is velocity, which is the rate at which displacement changes with respect to time.

Home > CBSE > CBSE Important Questions Class 12 Maths Chapter 7 – Integrals

CBSE Important Questions Class 12 Maths Chapter 7 – Integrals

Class 12 Maths Chapter 7 Integrals is part of the CBSE Class 12 Maths syllabus. Integration is the reverse process of differentiation. Students learn the standard integrals, integration by substitution, integration by partial fractions, integration by parts, definite integrals and their properties, and the Fundamental Theorem of Calculus.

These CBSE important questions for Class 12 Maths Chapter 7 cover the standard formulae, integration by substitution, integration by parts using the ILATE rule, partial fractions, definite integrals evaluated using the Fundamental Theorem, and the main properties of definite integrals. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Write the standard integration formulae for xⁿ, 1/x, e^x, sin x and cos x.

Answer: • ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, for n ≠ −1
• ∫(1/x) dx = log|x| + C
• ∫e^x dx = e^x + C
• ∫sin x dx = −cos x + C
• ∫cos x dx = sin x + C
• ∫sec²x dx = tan x + C
The constant C is called the constant of integration and must always be written for an indefinite integral.

Q2. Evaluate ∫(2x + 3)⁵ dx.

Answer: Use substitution. Let u = 2x + 3.
Then du = 2 dx, so dx = du/2.
∫u⁵ (du/2)
= ½ × u⁶/6 + C
= u⁶/12 + C
Putting back u = 2x + 3:
= (2x + 3)⁶/12 + C

Q3. State the rule for integration by parts and the ILATE order. Evaluate ∫x e^x dx.

Answer: Integration by parts:
∫u v dx = u ∫v dx − ∫[(du/dx) ∫v dx] dx
ILATE tells us which function to take as the first function u, in this order of preference:
Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.
For ∫x e^x dx: By ILATE, take u = x (algebraic) and v = e^x.
= x · e^x − ∫(1)(e^x) dx
= x e^x − e^x + C
= e^x(x − 1) + C

Q4. Evaluate ∫(sin x / (1 + cos x)) dx.

Answer: Use substitution. Let u = 1 + cos x.
Then du = −sin x dx, so sin x dx = −du.
∫(−du / u)
= −log|u| + C
Putting back u = 1 + cos x:
= −log|1 + cos x| + C

Q5. Evaluate the definite integral ∫₀¹ x² dx.

Answer: By the Fundamental Theorem of Calculus, if F is an antiderivative of f, then
∫ₐᵇ f(x) dx = F(b) − F(a)
Here the antiderivative of x² is x³/3.
∫₀¹ x² dx = [x³/3] from 0 to 1
= (1)³/3 − (0)³/3
= 1/3

Q6. Write the main properties of definite integrals.

Answer: • ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx
• ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx
• ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx
• ∫₀ᵇ f(x) dx = ∫₀ᵇ f(a − x) dx
• If f is an even function, ∫₋ₐₐ f(x) dx = 2∫₀ₐ f(x) dx
• If f is an odd function, ∫₋ₐₐ f(x) dx = 0

Q7. Evaluate ∫(1 / (x² − 9)) dx.

Answer: This matches the standard form ∫dx/(x² − a²) = (1/2a) log|(x − a)/(x + a)| + C.
Here a² = 9, so a = 3.
= (1/6) log|(x − 3)/(x + 3)| + C
So the answer is (1/6) log|(x − 3)/(x + 3)| + C.

Q8. Evaluate ∫((2x + 1) / ((x + 1)(x + 2))) dx using partial fractions.

Answer: Write the integrand as
(2x + 1)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2)
Multiplying through: 2x + 1 = A(x + 2) + B(x + 1)
Put x = −1: −1 = A(1), so A = −1
Put x = −2: −3 = B(−1), so B = 3
Now integrate each term:
∫[−1/(x + 1) + 3/(x + 2)] dx
= −log|x + 1| + 3 log|x + 2| + C

Q9. Evaluate ∫₀^π/2 sin x dx.

Answer: The antiderivative of sin x is −cos x.
∫₀^π/2 sin x dx = [−cos x] from 0 to π/2
= −cos(π/2) − (−cos 0)
= −0 + 1
= 1

Q10. Evaluate ∫log x dx.

Answer: Use integration by parts, taking log x as the first function (L comes before A in ILATE) and 1 as the second.
∫(log x)(1) dx
= log x · x − ∫(1/x)(x) dx
= x log x − ∫1 dx
= x log x − x + C
This can also be written as x(log x − 1) + C.

Q11. Evaluate ∫₋₁¹ x³ dx without doing the full calculation. Give a reason.

Answer: Check whether the function is odd or even.
f(x) = x³, and f(−x) = (−x)³ = −x³ = −f(x).
So f is an odd function.
For an odd function over a symmetric interval [−a, a]:
∫₋ₐₐ f(x) dx = 0
Therefore ∫₋₁¹ x³ dx = 0.
(The area above the x-axis exactly cancels the area below it.)

Q12. Evaluate ∫(x / (x² + 1)) dx.

Answer: Notice that the numerator is almost the derivative of the denominator.
Let u = x² + 1, so du = 2x dx, which gives x dx = du/2.
∫(1/u)(du/2)
= ½ log|u| + C
Putting back u = x² + 1:
= ½ log(x² + 1) + C
(We can drop the modulus because x² + 1 is always positive.)

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Q1.

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

Opt.

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

Ans.

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

Q2.

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

Opt.

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

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

$\begin{array}{l} [] \end{array}$ $\begin{array}{l} We have, \\ « \frac{\sin^{’ 1} x}{\sin^{’ 1} \sqrt{x} + \cos^{’ 1} \sqrt{x}} dx \\ « \frac{\sin^{’ 1} x}{\frac{}{2}} dx [µ \sin^{’ 1} \sqrt{x} + \cos^{’ 1} \sqrt{x} = \frac{}{2}] \\ = \frac{}{2} « \sin^{’ 1} x . dx [Using Integration by part] \\ = \sin^{’ 1} x « dx « [« dx] d (\sin^{’ 1} x) \\ = {xsin}^{’1} x « \frac{xdx}{\sqrt{1 ’ x^{2}}} \\ Put 1 x^{2} = t^{2} \\ On differentiating \\ 2 xdx = 2 tdt \\ xdx = tdt \\ = {xsin}^{’ 1} x + « \frac{tdt}{t} \\ = {xsin}^{’ 1} x + « dt \\ = {xsin}^{’ 1} x + t + c \\ = {xsin}^{’ 1} x + \sqrt{1 ’ x^{2}} + c \end{array}$

Q3.

$Evaluate_{0}^{\frac{}{2}} \log tanx dx .$

Opt.

$\begin{array}{l} Let I =_{0}^{\frac{}{2}} \log tanx dx . \dots. (1) \\ I =_{0}^{\frac{}{2}} \log \tan (\frac{}{2} x) dx \\ I = 0</m \end{array}$

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

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

CBSE Important Questions Class 12 Maths Chapter 7 – Integrals

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