Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.13 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.13

Choose the correct or the most suitable answer from given four alternatives.

Question 1.
If ∫ f(x) dx = g(x) + c. then ∫ f(x) g (x)dx
(1) ∫ (f(x)2 dx
(2) ∫ f(x) g(x) dx
(3) ∫ f'(x) g(x) dx
(4) ∫ (g(x))2 dx
(1) ∫ (f(x)2 dx

Explaination:
Given ∫ f (x) dx = g (x) + c
$$\frac{\mathrm{d}}{\mathrm{d} x}$$ ∫ f(x)dx = $$\frac{\mathrm{d}}{\mathrm{d} x}$$ (g(x) + c)
∫ $$\frac{\mathrm{d}}{\mathrm{d} x}$$ (f(x)) dx = g'(x)
∫ d(f(x)) = g'(x)
f(x) = g(x)
∴ ∫ f(x) g'(x) dx = ∫ f(x) f(x) dx
= ∫ [f(x)]2 dx

Question 2.
If $$\int \frac{3^{\frac{1}{x}}}{x^{2}}$$ dx = k$$\left(3^{\frac{1}{x}}\right)$$ + c, then the value of k is
(1) log 3
(2) – log 3
(3) $$-\frac{1}{\log 3}$$
(4) $$\frac{1}{\log 3}$$
(3) $$-\frac{1}{\log 3}$$

Explaination:

Question 3.
If ∫ f'(x) ex2 dx = (x – 1)ex2 + c, then f(x) is
(1) 2x3 – $$\frac{x^{2}}{2}$$ + x + c
(2) $$\frac{x^{3}}{2}$$ + 3x2 + 4x + c
(3) x3 + 4x2 + 6x + c
(4) $$\frac{2 x^{3}}{3}$$ – x2 + x + c
(4) $$\frac{2 x^{3}}{3}$$ – x2 + x + c

Explaination:
Given ∫ f'(x) ex2 dx = (x – 1)ex2 + c
Differentiating both sides with respect to x we have

Question 4.
The gradient (slope) of a curve at any point (x, y) is $$\frac{x^{2}-4}{x^{2}}$$. If the curve passes through the point (2, 7), then the equation of the curve is
(1) y = x + $$\frac{4}{x}$$ + 3
(2) y = x + $$\frac{4}{x}$$ + 4
(3)y = x2 + 3x + 4
(4) y = x2 – 3x + 6
(1) y = x + $$\frac{4}{x}$$ + 3

Explaination:

Given, this curve passes through the point (2, 7)
∴ 7 = 2 + $$\frac{4}{2}$$ + c
7 = 2 + 2 + c
c = 7 – 4 = 3
∴ The required equation is
y = x + $$\frac{4}{x}$$ + 3

Question 5.

(1) cot (xex) + c
(2) sec (xex) + c
(3) tan (xex) + c
(4) cos (xex) + c
(3) tan (xex) + c

Explaination:

= ∫sec2 t
= tan t + c
= tan (x ex) + c

Question 6.

(1) $$\sqrt{\tan x}+\mathbf{c}$$
(2) $$2 \sqrt{\tan x}+\mathbf{c}$$
(3) $$\frac{1}{2} \sqrt{\tan x}+c$$
(4) $$\frac{1}{4} \sqrt{\tan x}+c$$
(1) $$\sqrt{\tan x}+\mathbf{c}$$

Explaination:

Question 7.
∫sin3 dx is
(1)
(2)
(3)
(4)
(3)

Explaination:
∫sin3 dx
sin 3x = 3 sin x – 4 sin3 x
4 sin3 x = 3 sin x – sin 3x
sin3 x = $$\frac{1}{4}$$(3 sin x – sin 3x)

Question 8.

(1) x + c
(2) $$\frac{x^{3}}{3}$$ + c
(3) $$\frac{3}{x^{3}}$$ + c
(4) $$\frac{1}{x^{2}}$$ + c
(2) $$\frac{x^{3}}{3}$$ + c

Explaination:

Question 9.

(1) tan-1 (sin x) + c
(2) 2 sin-1 (tan x) + c
(3) tan-1 (cos x) + c
(4) sin-1 (tan x) + c
(4) sin-1 (tan x) + c

Explaination:

Question 10.

(1) x2 + c
(2) 2x2 + c
(3) $$\frac{x^{2}}{2}$$ + c
(4) – $$\frac{x^{2}}{2}$$ + c
(3) $$\frac{x^{2}}{2}$$ + c

Explaination:

Question 11.
∫23x+5 dx is
(1)
(2)
(3)
(4)
(4)

Explaination:
∫23x+5 dx
Put 3x + 5 = t
3 dx = dt
dx = $$\frac{1}{3}$$ dt

Question 12.

(1) $$\frac{1}{2}$$ sin 2x + c
(2) –$$\frac{1}{2}$$ sin 2x + c
(3) $$\frac{1}{2}$$ cos 2x + c
(4) –$$\frac{1}{2}$$ cos 2x + c
(2) –$$\frac{1}{2}$$ sin 2x + c

Explaination:

Question 13.

(1) ex tan-1 (x + 1) + c
(2) tan-1 (ex) + c
(3) ex $$\frac{\left(\tan ^{-1} x\right)^{2}}{2}$$ + c
(4) ex tan-1 x + c
(4) ex tan-1 x + c

Explaination:

Question 14.

(1) cot x + sin-1 x + c
(2) – cot x + tan-1x + c
(3) – tan x + cot-1 x + c
(4) – cot x – tan-1x + c
(4) – cot x – tan-1x + c

Explaination:

Question 15.
∫x2 cos x dx is
(1) x2 sin x + 2x cos x – 2 sin x + c
(2) x2 sin x – 2x cos x – 2 sin x + c
(3) – x2 sin x + 2x cos x + 2 sin x + c
(4) – x2 sin x – 2x cos x + 2 sin x + c
(1) x2 sin x + 2x cos x – 2 sin x + c

Explaination:
$$\int x^{2} \cos x d x$$
By Bernoullis formula dv = cosxdx
u = x2 v = sinx
u’ = 2x v1 = -cos x
u” = 2 v2 = -sinx
= uv – u’v1 + u”v2
= x2sin x + 2x cos x – 2 sin x + c

Question 16.

(1)
(2)
(3)
(4)
(1)

Explaination:

Question 17.
$$\int \frac{d x}{e^{x}-1}$$ is
(1) log |ex| – log |ex – 1| + c
(2) log |ex| + log |ex – 1| + c
(3) log |ex – 1| – log |ex| + c
(4) log |ex + 1| – log |ex| + c
(3) log |ex – 1| – log |ex| + c

Explaination:

Question 18.
∫e-4x cos x dx is
(1) $$\frac{e^{-4 x}}{17}$$ [4 cos x – sin x] + c
(2) $$\frac{e^{-4 x}}{17}$$ [- 4 cos x – sin x] + c
(3) $$\frac{e^{-4 x}}{17}$$ [4 cos x + sin x] + c
(4) $$\frac{e^{-4 x}}{17}$$ [- 4 cos x – sin x] + c
(2) $$\frac{e^{-4 x}}{17}$$ [- 4 cos x – sin x] + c

Explaination:

Question 19.

(1)
(2)
(3)
(4)
(4)

Explaination:

Question 20.
∫e-7x sin 5x dx is
(1) $$\frac{e^{-7 x}}{74}$$ [- 7 sin 5x – 5 cos 5x] + c
(2) $$\frac{e^{-7 x}}{74}$$ [7 sin 5x + 5 cos 5x] + c
(3) $$\frac{e^{-7 x}}{74}$$ [7 sin 5x – 5 cos 5x] + c
(4) $$\frac{e^{-7 x}}{74}$$ [- 7 sin 5x + 5 cos 5x] + c
(1) $$\frac{e^{-7 x}}{74}$$ [- 7 sin 5x – 5 cos 5x] + c

Explaination:

Question 21.
∫x2 ex/2 dx is
(1)
(2)
(3)
(4)
(3)

Explaination:

Question 22.

(1)
(2)
(3)
(4)
(4)

Explaination:

Question 23.

(1)
(2)
(3)
(4)
(3)

Explaination:

Question 24.
∫sin √x dx is
(1) 2(- √x cos √x + sin √x) + c
(2) 2(- √x cos √x + sin √x) + c
(3) 2(- √x sin √x – cos √x) + c
(4) 2(- √x sin √x + cos √x) + c
(1) 2(- √x cos √x + sin √x) + c

Explaination:

Question 25.

(1)
(2)
(3)
(4)