Question 1

The value of the integral _ 0 ^ 2 x(x^2+x+1) ( x+1 )( x^4+x^2+1 ) \, dx is equal to:

Accepted Answer

Let I = _ 0 ^ 2 x(x^2 + x + 1) x + 1 \, x^4 + x^2 + 1 \, dx. Using the identity x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1): I = _ 0 ^ 2 x \, x^2 + x + 1 x + 1 \, (x^2 + x + 1)(x^2 - x + 1) \, dx Cancelling x^2 + x + 1 : I = _ 0 ^ 2 x (x + 1)(x^2 - x + 1) \, dx Using (x + 1)(x^2 - x + 1) = x^3 + 1: I = _ 0 ^ 2 x x^3 + 1 \, dx Substitution: Let t = x^ 3/2 , so dt = 3 2 x^ 1/2 \, dx x \, dx = 2 3 \, dt. Also, t^2 = x^3, so x^3 + 1 = t^2 + 1. Changing limits: x = 0 t = 0 x = 2 t = 2^ 3/2 = 2 2 There

Question 2

If _ /6 ^ /4 ( (x- 3 ) (x+ 3 )+1 )dx = _e( 3 -1), then 9 ^2 is equal to ________.

Accepted Answer

Given integral is I = _ /6 ^ /4 ( (x- 3 ) (x+ 3 )+1 )dx Using the identity A B + 1 = (A-B) A B , the integrand simplifies to: f(x) = ( (x- 3 ) - (x+ 3 ) ) (x- 3 ) (x+ 3 ) f(x) = (- 2 3 ) ^2 x - ^2 ( 3 ) f(x) = -1/2 ^2 x - 3/4 = 2 3 - 4 ^2 x Dividing the numerator and the denominator by ^2 x: f(x) = 2 ^2 x 3 ^2 x - 4 ^2 x = 2 ^2 x 3(1+ ^2 x) - 4 ^2 x = 2 ^2 x 3 - ^2 x The integral becomes: I = _ /6 ^ /4 2 ^2 x 3 - ^2 x dx Substituting x = t ^2 x dx = dt. When x = 6 , t = 1 3 . When x = 4 , t = 1.

Question 3

Let A = bmatrix 1 & 3 & -1 \ 2 & 1 & \ 0 & 1 & -1 bmatrix be a singular matrix. Let f(x) = _0^x (t^2 + 2t + 3)\,dt, x [1, ]. If M and m are respectively the maximum and the minimum values of f in [1

Accepted Answer

Since A is a singular matrix, |A| = 0. |A| = 1(-1 - ) - 3(-2 - 0) - 1(2 - 0) = 0 -1 - + 6 - 2 = 0 = 3 The function is f(x) = _0^x (t^2 + 2t + 3)\,dt = x^3 3 + x^2 + 3x Differentiating with respect to x, we get f'(x) = x^2 + 2x + 3 = (x+1)^2 + 2 > 0 for all x. Thus, f(x) is a strictly increasing function on the interval [1, 3]. The maximum value M occurs at x = 3: M = f(3) = 3^3 3 + 3^2 + 3(3) = 9 + 9 + 9 = 27 The minimum value m occurs at x = 1: m = f(1) = 1^3 3 + 1^2 + 3(1) = 13 3 We need to fi

Question 4

Let f: R R be such that f(xy) = f(x)f(y), for all x, y R and f(0) 0. Let g: [1, ) R be a differentiable function such that x^2 g(x) = _1^x (t^2 f(t) - tg(t))\,dt. Then g(2) is equal to :

Accepted Answer

Given f(xy) = f(x)f(y) for all x, y R and f(0) 0. Substituting y = 0, we get f(0) = f(x)f(0). Since f(0) 0, dividing by f(0) gives f(x) = 1 for all x R . The given integral equation is: x^2 g(x) = _1^x (t^2 f(t) - tg(t))\,dt Substituting f(t) = 1: x^2 g(x) = _1^x (t^2 - tg(t))\,dt Differentiating both sides with respect to x using Leibniz's rule: 2x g(x) + x^2 g'(x) = x^2 - xg(x) x^2 g'(x) + 3xg(x) = x^2 Dividing by x (since x 1): x g'(x) + 3g(x) = x g'(x) + 3 x g(x) = 1 This is a linear differe

Question 5

The value of the integral _ -1 ^ 1 ( x^3 + |x| + 1 x^2 + 2|x| + 1 ) dx is equal to :

Accepted Answer

Let I = _ -1 ^ 1 ( x^3 + |x| + 1 x^2 + 2|x| + 1 ) dx We can split the integral into two parts: I = _ -1 ^ 1 x^3 x^2 + 2|x| + 1 dx + _ -1 ^ 1 |x| + 1 x^2 + 2|x| + 1 dx The first integrand f(x) = x^3 x^2 + 2|x| + 1 is an odd function since f(-x) = -f(x). Therefore, its integral over the symmetric interval [-1, 1] is zero. The second integrand g(x) = |x| + 1 x^2 + 2|x| + 1 is an even function since g(-x) = g(x). Therefore, its integral over [-1, 1] is twice the integral over [0, 1]. I = 0 + 2 _ 0 ^

Question 6

The value of the integral _ - /4 ^ /4 ( 32 ^4 x 1 + e^ x )dx is:

Accepted Answer

Let I = _ - /4 ^ /4 32 ^4 x 1 + e^ x dx Using the definite integral property _ -a ^ a f(x) dx = _ 0 ^ a (f(x) + f(-x)) dx, we get: I = _ 0 ^ /4 ( 32 ^4 x 1 + e^ x + 32 ^4(-x) 1 + e^ (-x) ) dx I = _ 0 ^ /4 ( 32 ^4 x 1 + e^ x + 32 ^4 x 1 + e^ - x ) dx I = _ 0 ^ /4 ( 32 ^4 x 1 + e^ x + 32 ^4 x e^ x e^ x + 1 ) dx I = _ 0 ^ /4 32 ^4 x ( 1 + e^ x 1 + e^ x ) dx I = _ 0 ^ /4 32 ^4 x dx Using the trigonometric identity ^2 x = 1 + 2x 2 , we can write: ^4 x = ( 1 + 2x 2 )^2 = 1 4 (1 + 2 2x + ^2 2x) ^4 x =

Question 7

Let (2^ 1-a + 2^ 1+a ), f(a), (3^a + 3^ -a ) be in A.P. and be the minimum value of f(a). Then the value of the integral _ _e( -1) ^ _e( ) dx (e^ 2x - e^ -2x ) is :

Accepted Answer

Since (2^ 1-a + 2^ 1+a ), f(a), and (3^a + 3^ -a ) are in A.P., we have: 2f(a) = 2^ 1-a + 2^ 1+a + 3^a + 3^ -a f(a) = (2^ -a + 2^a) + 1 2 (3^a + 3^ -a ) Using the AM-GM inequality, x + 1 x 2 for x > 0. 2^a + 2^ -a 2 and 3^a + 3^ -a 2. The minimum value of f(a) occurs when a = 0. = f(0) = (2^0 + 2^0) + 1 2 (3^0 + 3^0) = 2 + 1 = 3. Now, we evaluate the integral: I = _ _e 2 ^ _e 3 dx e^ 2x - e^ -2x = _ _e 2 ^ _e 3 e^ 2x e^ 4x - 1 dx Let e^ 2x = t 2e^ 2x dx = dt. When x = _e 2, t = e^ 2 _e 2 = 4. Wh

Question 8

Let f: [1, ) R be a differentiable function defined as f(x) = _1^x f(t)\,dt + (1-x)( _e x - 1) + e. Then the value of f(f(1)) is :

Accepted Answer

Given f(x) = _1^x f(t)\,dt + (1-x)( _e x - 1) + e Substituting x = 1 in the given equation: f(1) = _1^1 f(t)\,dt + (1-1)( _e 1 - 1) + e f(1) = 0 + 0 + e = e Differentiating the given equation with respect to x using the Leibniz rule: f'(x) = f(x) + d dx [(1-x)( _e x - 1)] f'(x) = f(x) + (-1)( _e x - 1) + (1-x) ( 1 x ) f'(x) = f(x) - _e x + 1 + 1 x - 1 f'(x) - f(x) = 1 x - _e x This is a linear differential equation of the form dy dx + Py = Q, where P = -1 and Q = 1 x - _e x. Integrating Factor (

Question 9

The value of the integral _0^ _e(x) x^2 + 4 \,dx is:

Accepted Answer

Let I = _0^ _e(x) x^2 + 4 \,dx Substitute x = 2 , which gives dx = 2 ^2 \,d . The limits of integration change from x = 0 = 0 to x = 2 . I = _0^ /2 _e(2 ) 4 ^2 + 4 (2 ^2 ) \,d I = _0^ /2 _e(2 ) 4 ^2 (2 ^2 ) \,d I = 1 2 _0^ /2 _e(2 ) \,d I = 1 2 _0^ /2 ( _e 2 + _e( )) \,d I = 1 2 _e 2 _0^ /2 1 \,d + 1 2 _0^ /2 _e( ) \,d Let I_1 = _0^ /2 _e( ) \,d Using the property _0^a f(x)\,dx = _0^a f(a-x)\,dx, we get: I_1 = _0^ /2 _e ( ( 2 - ) ) \,d = _0^ /2 _e( ) \,d I_1 = _0^ /2 _e ( 1 ) \,d = - _0^ /2 _e(

Question 10

The value of the integral _ /6 ^ /3 ( 4 - ^2 x ^4 x ) dx is:

Accepted Answer

The given integral is I = _ /6 ^ /3 ( 4 - ^2 x ^4 x ) dx. We can rewrite the integrand in terms of x and x. Using the identity ^2 x = 1 + ^2 x = 1 + 1 ^2 x , we get: I = _ /6 ^ /3 (4 - 1 - 1 ^2 x ) ^4 x \, dx I = _ /6 ^ /3 (3 - 1 ^2 x ) (1 + ^2 x) ^2 x \, dx Substitute t = x, which gives dt = ^2 x \, dx. The limits of integration change as follows: When x = 6 , t = 1 3 . When x = 3 , t = 3 . The integral becomes: I = _ 1/ 3 ^ 3 (3 - 1 t^2 ) (1 + t^2) dt Expanding the integrand, we get: I = _ 1/

Question 11

The integral _ 0 ^ 1 ^ -1 (1+x+x^2)dx is equal to:

Accepted Answer

I = _ 0 ^ 1 ^ -1 (1+x+x^2)dx I = _ 0 ^ 1 ^ -1 ( 1 1+x+x^2 )dx I = _ 0 ^ 1 ^ -1 ( (x+1)-x 1+(x+1)x )dx I = _ 0 ^ 1 ( ^ -1 (x+1)- ^ -1 x)dx I = _ 0 ^ 1 ^ -1 (x+1)dx - _ 0 ^ 1 ^ -1 x dx Substituting x+1 = t in the first integral: I = _ 1 ^ 2 ^ -1 x dx - _ 0 ^ 1 ^ -1 x dx Using integration by parts, ^ -1 x dx = x ^ -1 x - 1 2 _e(1+x^2) I = [x ^ -1 x - 1 2 _e(1+x^2) ]_ 1 ^ 2 - [x ^ -1 x - 1 2 _e(1+x^2) ]_ 0 ^ 1 I = (2 ^ -1 2 - 1 2 _e 5 - ( 4 - 1 2 _e 2 ) ) - ( 4 - 1 2 _e 2 - 0 ) I = 2 ^ -1 2 - 1 2 _e

Question 12

Let f be a twice differentiable function such that f(x)= _ 0 ^ x (t-x)dt- _ 0 ^ x f(t) t\,dt, x (- 2 , 2 ). Then f'' ( 6 )+12f' (- 6 )+f ( 6 ) is equal to ______

Accepted Answer

Given the function: f(x) = _ 0 ^ x (t-x) dt - _ 0 ^ x f(t) t \, dt First, we simplify the first integral. Let u = x - t, then du = -dt. When t = 0, u = x; when t = x, u = 0. _ 0 ^ x (t-x) dt = _ x ^ 0 (-u) (-du) = _ x ^ 0 u \, du = - _ 0 ^ x u \, du = -[ ( u)]_ 0 ^ x = - ( x) = ( x) So, the equation becomes: f(x) = ( x) - _ 0 ^ x f(t) t \, dt Differentiating both sides with respect to x using the Leibniz rule: f'(x) = - x x - f(x) x f'(x) + f(x) x = - x This is a linear first-order differential

Question 13

Let _ -2 ^ 2 (| x| + [x x])\,dx = 2(3 - 2) + , where [ ] is the greatest integer function. Then ( 2 ) equals:

Accepted Answer

Let I = _ -2 ^ 2 (| x| + [x x] ) dx = I_1 + I_2, where I_1 = _ -2 ^ 2 | x|\, dx and I_2 = _ -2 ^ 2 [x x]\, dx. Evaluating I_1: Since | x| is even, I_1 = 2 _ 0 ^ 2 | x|\, dx. For x [0, 2], 2 I_1 = 2 _ 0 ^ 2 x\, dx = 2 [- x ]_0^2 = 2(1 - 2) = 2 - 2 2 Evaluating I_2: Let g(x) = x x. Then g(-x) = (-x) (-x) = x x = g(x), so g(x) is even, and hence [x x] is also even. I_2 = 2 _ 0 ^ 2 [x x]\, dx Behaviour of y = x x on [0, 2]: At x = 0, y = 0. At x = 2, y = 2 2 1.818. y' = x + x x > 0 on [0, 2], so y i

Question 14

The value of _0^ 20 ( ^4 x + ^4 x) \, dx is equal to:

Accepted Answer

Let I = _0^ 20 ( ^4 x + ^4 x) \, dx Using the algebraic identity a^2 + b^2 = (a+b)^2 - 2ab, the integrand can be simplified: ^4 x + ^4 x = ( ^2 x + ^2 x)^2 - 2 ^2 x ^2 x Since ^2 x + ^2 x = 1 and 2 x x = 2x: ^4 x + ^4 x = 1 - 1 2 (2 x x)^2 = 1 - 1 2 ^2 2x Using the half-angle formula ^2 = 1 - 2 2 : ^4 x + ^4 x = 1 - 1 2 ( 1 - 4x 2 ) = 1 - 1 4 + 1 4 4x = 3 4 + 1 4 4x Substituting this back into the integral: I = _0^ 20 ( 3 4 + 1 4 4x ) \, dx I = [ 3 4 x + 1 16 4x ]_0^ 20 Evaluating the limits: I

Question 15

Let [ ] denote the greatest integer function. Then the value of _0^3 ( e^x + e^ -x [x]! ) dx is :

Accepted Answer

The given integral can be split at integer values of x because of the greatest integer function [x]. I = _0^3 ( e^x + e^ -x [x]! ) dx I = _0^1 e^x + e^ -x 0! dx + _1^2 e^x + e^ -x 1! dx + _2^3 e^x + e^ -x 2! dx Since 0! = 1, 1! = 1, and 2! = 2, we get: I = _0^2 (e^x + e^ -x ) dx + 1 2 _2^3 (e^x + e^ -x ) dx Integrating the terms: I = [ e^x - e^ -x ]_0^2 + 1 2 [ e^x - e^ -x ]_2^3 I = (e^2 - e^ -2 - (1 - 1)) + 1 2 (e^3 - e^ -3 - (e^2 - e^ -2 )) I = e^2 - e^ -2 + 1 2 e^3 - 1 2 e^ -3 - 1 2 e^2 + 1 2

Question 16

If = _0^ 2 3 _2(x^2 + 4)\,dx + _2^4 2^x - 4 \,dx, then ^2 is equal to _______.

Accepted Answer

Let f(x) = _2(x^2 + 4) for x 0. To find the inverse function f^ -1 (x), we set y = _2(x^2 + 4) and solve for x: 2^y = x^2 + 4 x^2 = 2^y - 4 x = 2^y - 4 Thus, f^ -1 (x) = 2^x - 4 . The given expression is = _0^ 2 3 f(x)\,dx + _2^4 f^ -1 (x)\,dx. We observe the limits of the first integral and evaluate the function at these points: f(0) = _2(0 + 4) = 2 f(2 3 ) = _2((2 3 )^2 + 4) = _2(12 + 4) = _2(16) = 4 These exactly match the limits of the second integral. Using the standard property of definite

Question 17

Let [ ] denote the greatest integer function. Then _ - 2 ^ 2 ( 12(3+[x]) 3+[ x]+[ x] ) d x is equal to :

Accepted Answer

I = _ - /2 ^ /2 12(3 + [x]) \, dx 3 + [ x] + [ x] I = _ - /2 ^ -1 12(1) \, dx 2 + _ -1 ^ 0 12(2) \, dx 2 + _ 0 ^ 1 12(3) \, dx 3 + _ 1 ^ /2 12(4) \, dx 3 I = 6 ( 2 - 1 ) + 12(0 + 1) + 12(1 - 0) + 16 ( 2 - 1 ) I = 3 - 6 + 12 + 12 + 8 - 16 I = 11 + 2

Question 18

Let f be a polynomial function such that f (x^ 2 +1 )=x^ 4 +5 x^ 2 +2, for all x R . Then _ 0 ^ 3 f(x) d x is equal to

Accepted Answer

Let u = x^2 + 1, then x^2 = u - 1. Substituting into the given condition: f(u) = (u-1)^2 + 5(u-1) + 2 = u^2 - 2u + 1 + 5u - 5 + 2 = u^2 + 3u - 2. Therefore, f(x) = x^2 + 3x - 2. Calculating the integral: _0^3 f(x)dx = _0^3 (x^2 + 3x - 2)dx = [ x^3 3 + 3x^2 2 - 2x ]_0^3 = 9 + 27 2 - 6 = 33 2 .

Question 19

If ( 1-5 ^ 2 x ^ 5 x ^ 2 x ) d x=f(x)+ C , where C is the constant of integration, then f ( 6 )-f ( 4 ) is equal to

Accepted Answer

1 - 5 ^2 x ^5 x ^2 x dx = dx ^5 x ^2 x - 5 dx ^5 x = ^2 x \, dx ^5 x - 5 dx ^5 x By IBP: = x ^5 x - (- 5 ^6 x ) x x \, dx - 5 dx ^5 x = x ^5 x + C So f(x) = x ^5 x f ( 6 ) - f ( 4 ) = 2^5 3 - ( 2 )^5 = 32 3 - 4 2 = 4 3 (8 - 6 )

Question 20

The value of _ r=1 ^ 20 ( | ( _ 0 ^ r x| x| d x ) | ) is \_\_\_\_

Accepted Answer

Let I_r = _ 0 ^ r x| x| \, dx .....(1) Apply King Property = _ 0 ^ r (r - x)| x| \, dx .....(2) By (1) + (2) 2I_r = _ 0 ^ r r| x| \, dx I_r = r 2 _ 0 ^ r | x| \, dx I_1 = 1 2 _ 0 ^ 1 | x| \, dx = 1 2 _ 0 ^ | t| \, dt = 1 2 (2) I_2 = 2 2 _ 0 ^ 2 | x| \, dx = 2 2 _ 0 ^ 2 | t| \, dt = 2 2 (4) S = 1 2 2 + 2 2 4 + 3 2 6 + + 20 2 (2 20) = 1 + 2 + 3 + + 20 = 20 21 2 = 210

Definite Integration — Previous Year Questions & Solutions