Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.12 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.12

Choose the most suitable answer from the given four alternatives:

Question 1.
∫$$\frac { 1 }{x^3}$$ dx is
(a) $$\frac { -3 }{x^2}$$ + c
(b) $$\frac { -1 }{2x^2}$$ + c
(c) $$\frac { -1 }{3x^2}$$ + c
(d) $$\frac { -2 }{x^2}$$ + c
Solution:
(b) $$\frac { -1 }{2x^2}$$ + c
Hint:
∫$$\frac { 1 }{x^3}$$ dx = ∫x-3 dx = [ $$\frac { x^{-3+1} }{-3+1}$$ ] + c
= ($$\frac { x^{-2} }{-2}$$) + c = $$\frac { -1 }{2x^2}$$ + c

Question 2.
∫2x dx is
(a) 2x log 2 + c
(b) 2x + c
(c) $$\frac { 2^x }{log 2}$$ + c
(d) $$\frac { log 2 }{2^x}$$ + c
Solution:
(c) $$\frac { 2^x }{log 2}$$ + c
Hint:
∫2x dx = ∫ax dx = $$\frac { a^x }{log a}$$ + c

Question 3.
∫$$\frac { sin 2x }{2 sin x}$$ dx is
(a) sin x + c
(b) $$\frac { 1 }{2}$$ sin x + c
(c) cos x + c
(d) $$\frac { 1 }{2}$$ cos x + c
Solution:
(a) sin x + c
Hint:
∫$$\frac { sin 2x }{2 sin x}$$ dx = ∫$$\frac { 2sin x cos x }{2 sin x}$$ dx
= ∫cos x dx
= sin x + c

Question 4.
∫$$\frac { sin 5x-sin x }{cos 3x}$$ dx is
(a) -cos 2x + c
(b) -cos 2x – c
(c) –$$\frac { 1 }{4}$$ cos 2x + c
(d) -4 cos 2x + c
Solution:
(a) -cos 2x + c
Hint:

Question 5.
∫$$\frac { log x}{x}$$ dx, x > 0 is
(a) $$\frac { 1 }{2}$$ (log x)² + c
(b) –$$\frac { 1 }{2}$$ (log x)²
(c) $$\frac { 2 }{x^2}$$ + c
(d) $$\frac { 2 }{x^2}$$ – c
Solution:
(a) $$\frac { 1 }{2}$$ (log x)² + c
Hint:
∫$$\frac { log x}{x}$$ dx, x > 0
∫ tdt = [ $$\frac { t^2 }{2}$$ ] + c
= $$\frac { (log x)^2 }{2}$$ + c
let t = log x
$$\frac { dt }{dx}$$ = $$\frac { 1 }{x}$$
dt = $$\frac { 1 }{x}$$ dx

Question 6.
∫$$\frac { e^x }{\sqrt{1+e^x}}$$ dx is
(a) $$\frac { e^x }{\sqrt{1+e^x}}$$ + c
(b) 2$$\sqrt{1+e^x}$$ + c
(c) $$\sqrt{1+e^x}$$ + c
(d) ex$$\sqrt{1+e^x}$$ + c
Solution:
(b) 2$$\sqrt{1+e^x}$$ + c
Hint:

Question 7.
∫$$\sqrt { e^x}$$ dx is
(a) $$\sqrt { e^x}$$ + c
(b) 2$$\sqrt { e^x}$$ + c
(c) $$\frac { 1 }{2}$$ $$\sqrt { e^x}$$ + c
(d) $$\frac { 1 }{2\sqrt { e^x}}$$ + c
Solution:
(b) 2$$\sqrt { e^x}$$ + c
Hint:
∫$$\sqrt { e^x}$$ dx
= ∫$$\sqrt { e^x}$$ dx = ∫(ex)1/2 dx = ∫ ex/2 dx
= $$\frac { e^{x/2} }{1/2}$$ + c = 2ex/2 + c
= 2(ex)1/2 + c = 2$$\sqrt { e^x}$$ + c

Question 8.
∫e2x [2x² + 2x] dx
(a) e2x x² + c
(b) xe2x + c
(c) 2x²e² + c
(d) $$\frac { x^2e^x }{2}$$ + c
Solution:
(a) e2x x² + c
Hint:
∫e2x (2x² + 2x) dx
Let f(x) = x²; f'(x) = 2x and a = 2
= ∫eax [af(x),+ f ’(x)] = eax f(x) + c
= ∫e2x (2x² + 2x) dx = e2x (x²) + c

Question 9.
$$\frac { e^x }{e^x+1}$$ dx is
(a) log |$$\frac { e^x }{e^x+1}$$| + c
(b) log |$$\frac { e^x+1 }{e^x}$$| + c
(c) log |ex| + c
(d) log |ex + 1| + c
Solution:
(d) log |ex + 1| + c
Hint:
∫$$\frac { e^x }{e^x+1}$$ dx
= ∫$$\frac { dt }{t}$$
= log |t| + c
= log |ex + 1| + c
take t = ex + 1
$$\frac { dt }{dx}$$ = ex
dt = ex dx

Question 10.
∫$$\frac { 9 }{x-3}-\frac { 1 }{x+1}$$ dx is
(a) log |x – 3| – log|x + 1| + c
(b) log|x – 3| + log|x + 1| + c
(c) 9 log |x – 3| – log |x + 1| + c
(d) 9 log |x – 3| + log |x + 1| + c
Solution:
(c) 9 log |x – 3| – log |x + 1| + c
Hint:

Question 11.
∫$$\frac { 2x^3 }{4+x^4}$$ dx is
(a) log |4 + x4| + c
(b) $$\frac { 1 }{2}$$ log |4 + x4| + c
(c) $$\frac { 1 }{2}$$ log |4 + x4| + c
(d) log |$$\frac { 2x^3 }{4+x^4}$$ + c
Solution:
(b) $$\frac { 1 }{2}$$ log |4 + x4| + c
Hint:

Question 12.
∫$$\frac { dx }{\sqrt{x^2-36}}$$ is
(a) $$\sqrt{x^2-36}$$ + c
(b) log |x + $$\sqrt{x^2-36}$$| + c
(c) log |x – $$\sqrt{x^2-36}$$| + c
(d) log |x² + $$\sqrt{x^2-36}$$| + c
Solution:
(b) log |x + $$\sqrt{x^2-36}$$| + c
Hint:

Question 13.
∫$$\frac { 2x+3 }{\sqrt{x^2+3x+2}}$$ dx is
(a) $$\sqrt{x^2+3x+2}$$ + c
(b) 2$$\sqrt{x^2+3x+2}$$ + c
(c) $$\sqrt{x^2+3x+2}$$ + c
(d) $$\frac { 2 }{3}$$ (x² + 3x + 2) + c
Solution:
(b) 2$$\sqrt{x^2+3x+2}$$ + c
Hint:

Question 14.
$$\int_{0}^{4}$$ (2x + 1) dx is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:
$$\int_{0}^{4}$$ (2x + 1) dx
= [2($$\frac { x^2 }{2}$$) + x]$$_{0}^{1}$$ = [x² + x]$$_{0}^{1}$$
= [(1)² + (1)] – [0] = 2

Question 15.
$$\int_{2}^{4}$$ $$\frac { dx }{x}$$ is
(a) log 4
(b) 0
(c) log 2
(d) log 8
Solution:
(c) log 2
Hint:
$$\int_{2}^{4}$$ $$\frac { dx }{x}$$
$$\int_{2}^{4}$$ $$\frac { dx }{x}$$ = [log |x|]$$_{0}^{1}$$ = log |4| – log |2|
= log[ $$\frac { 4}{2}$$ ] = log 2

Question 16.
$$\int_{0}^{∞}$$ e-2x dx is
(a) 0
(b) 1
(c) 2
(d) $$\frac { 1 }{2}$$
Solution:
(d) $$\frac { 1 }{2}$$
Hint:

Question 17.
$$\int_{-1}^{1}$$ x³ ex4 dx is
(a) 1
(b) 2$$\int_{0}^{1}$$ x³ ex4
(c) 0
(d) ex4
Solution:
(c) 0
Hint:
$$\int_{-1}^{1}$$ x³ ex4 dx
Let f (x) = x³ex4
f(-x) = (-x)² e(-x)4
= -x² ex4
f(-x) = -f(x)
⇒ f(x) is an odd function
∴ $$\int_{-1}^{1}$$ x³ ex4 dx = 0

Question 18.
If f(x) is a continuous function and a < c < b, then $$\int_{a}^{c}$$ f(x) dx + $$\int_{c}^{b}$$ f(x) dx is
(a) $$\int_{a}^{b}$$ f(x) dx – $$\int_{a}^{c}$$ f(x) dx
(b) $$\int_{a}^{c}$$ f(x) dx – $$\int_{a}^{b}$$ f(x) dx
(c) $$\int_{a}^{b}$$ f(x) dx
(d) 0
Solution:
(c) $$\int_{a}^{b}$$ f(x) dx

Question 19.
The value of $$\int_{-π/2}^{π/2}$$ cos x dx is
(a) 0
(b) 2
(c) 1
(d) 4
Solution:
(b) 2
Hint:
$$\int_{-π/2}^{π/2}$$ cos x dx
Let f(x) = cos x
f(-x) = cos (-x) = cos (x) = f(x)
∴ f(x) is an even function
$$\int_{-π/2}^{π/2}$$ cos x dx = 2 × $$\int_{0}^{π/2}$$ cos x dx
= 2 × [sin x]$$_{0}^{-π/2}$$ = 2 [sin π/2 – sin 0]
= 2 [1 – 0] = 2

Question 20.
$$\int_{-π/2}^{π/2}$$ $$\sqrt {x^4(1-x)^2}$$ dx
(a) $$\frac { 1 }{12}$$
(b) $$\frac { -7 }{12}$$
(c) $$\frac { 7 }{12}$$
(d) $$\frac { -1 }{12}$$
Solution:
(a) $$\frac { 1 }{12}$$
Hint:

Question 21.
If $$\int_{0}^{1}$$ f(x) dx = 1, $$\int_{0}^{1}$$ x f(x) dx = a and $$\int_{0}^{1}$$ x² f(x) dx = a², then $$\int_{0}^{1}$$ (a – x)² f(x) dx is
(a) 4a²
(b) 0
(c) a²
(d) 1
Solution:
(b) 0
Hint:
$$\int_{0}^{1}$$ (a – x)² f(x) dx
= $$\int_{0}^{1}$$ [a² +x² – 2ax] f(x) dx
= $$\int_{0}^{1}$$ a² + f (x) dx + $$\int_{0}^{1}$$ x² f (x) dx – 2a$$\int_{0}^{1}$$ x f(x) dx
= a²(1) + a² – 2a(a) – 2a² – 2a² = 0

Question 22.
The value of $$\int_{2}^{3}$$ f(5 – x) dx – $$\int_{2}^{3}$$ f(x) dx is
(a) 1
(b) 0
(c) -1
(d) 5
Solution:
(b) 0
Hint:
$$\int_{2}^{3}$$ f(5 – x) dx – $$\int_{2}^{3}$$ f(x) dx
Using the property
= $$\int_{2}^{3}$$ f(x) dx = $$\int_{a}^{b}$$ f(a + b – x) dx
= $$\int_{2}^{3}$$ f (5 – x) – $$\int_{2}^{3}$$ f (5 – x) dx
= 0

Question 23.
$$\int_{0}^{4}$$ (√x + $$\frac { 1 }{√x}$$), dx is
(a) $$\frac { 20 }{3}$$
(b) $$\frac { 21 }{3}$$
(c) $$\frac { 28 }{3}$$
(d) $$\frac { 1 }{3}$$
Solution:
(c) $$\frac { 28 }{3}$$
Hint:

Question 24.
$$\int_{0}^{π/3}$$ tan x dx is
(a) log 2
(b) 0
(c) log √2
(d) 2 log 2
Solution:
(a) log 2
Hint:
$$\int_{0}^{π/3}$$ tan x dx
= ∫tan x dx
= ∫$$\frac { sin x }{cos x}$$ dx
= -∫$$\frac { -sin x }{cos x}$$ dx
= -log |cos x| + c
= log sec x + c
= [log (sec x)]$$_{0}^{π/3}$$
= log [(sec π/3) – log (sec 0)]
= log (2) – log (1)
= log 2 – (0) = log 2

Question 25.
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8
(a) 5040
(b) 5400
(c) 4500
(d) 5540
Solution:
(a) 5040
Hint:
$$\Upsilon$$ (8) = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Question 26.
Γ(n) is
(a) (n – 1)!
(b) n!
(c) n Γ (n)
(d) (n – 1) Γ(n)
Solution:
(a) (n – 1)!
Hint:
Γ(n) = Γ(n – 1) + 1 = (n – 1)!

Question 27.
Γ(1) is
(a) 0
(b) 1
(c) n
(d) n!
Solution:
(b) 1
Hint:
$$\Upsilon$$ (1) = 0! = 1

Question 28.
If n > 0, then Γ(n) is
(a) $$\int_{0}^{1}$$ e-x xn-1 dx
(b) $$\int_{0}^{1}$$ e-x xⁿ dx
(c) $$\int_{0}^{∞}$$ ex x-n dx
(d) $$\int_{0}^{∞}$$ e-x xn-1 dx
Solution:
(d) $$\int_{0}^{∞}$$ e-x xn-1 dx

Question 29.
Γ($$\frac { 3 }{2}$$)
(a) √π
(b) $$\frac { √π }{2}$$
(c) 2√π
(d) $$\frac { 3 }{2}$$
Solution:
(b) $$\frac { √π }{2}$$
Hint:
$$\Upsilon$$ (3/2) = $$\frac { 2 }{2}$$ $$\Upsilon$$ [ $$\frac { 3 }{2}$$ ]
= $$\frac { 3 }{2}$$ √π

Question 30.
$$\int_{0}^{∞}$$ x4 e-x dx is
(a) 12
(b) 4
(c) 4!
(d) 64
Solution:
(b) $$\frac { √π }{2}$$
Hint:
$$\int_{0}^{∞}$$ x4 e-x dx
= ∫xⁿ e-ax dx = $$\frac { n! }{a{n+1}}$$
= $$\frac { 4! }{(1)^{n+1}}$$
= $$\frac { 4! }{(1)^5}$$
= 4!