Question 1

Passage: Let f(x)= x and g(x)-f(x)=f(4-x). Question: What is _0^4 f(x) g(x) \,dx equal to ?

Accepted Answer

I = _0^4 f(x) g(x) \,dx From the given equation, g(x) = f(x) + f(4-x) Using the property _0^a h(x) \,dx = _0^a h(a-x) \,dx: I = _0^4 f(4-x) g(4-x) \,dx Since g(4-x) = f(4-x) + f(x) = g(x): I = _0^4 f(4-x) g(x) \,dx Adding the two equations for I: 2I = _0^4 f(x) + f(4-x) g(x) \,dx 2I = _0^4 g(x) g(x) \,dx = _0^4 1 \,dx 2I = 4 I = 2 Answer: 2

Question 2

Passage: Let _0^ 2 a x+b x (a+b)( x+ x) \,dx=k. Question: What is the value of k ?

Accepted Answer

Let I = _0^ 2 a x+b x (a+b)( x+ x) \,dx Using the property _0^a f(x)\,dx = _0^a f(a-x)\,dx, we get: I = _0^ 2 a ( 2 -x )+b ( 2 -x ) (a+b) ( ( 2 -x )+ ( 2 -x ) ) \,dx I = _0^ 2 a x+b x (a+b)( x+ x) \,dx Adding the two equations for I, we get: 2I = _0^ 2 a x+b x+a x+b x (a+b)( x+ x) \,dx 2I = _0^ 2 (a+b)( x+ x) (a+b)( x+ x) \,dx 2I = _0^ 2 1 \,dx 2I = 2 I = 4 Thus, k = 4 Answer: 4

Question 3

Passage: Let f(x)= x and g(x)-f(x)=f(4-x). Question: What is _0^4 f(4-x) g(4-x) \,dx equal to ?

Accepted Answer

Given g(x) - f(x) = f(4-x) g(x) = f(x) + f(4-x) Replacing x with 4-x, we get: g(4-x) = f(4-x) + f(x) = g(x) Let I = _0^4 f(4-x) g(4-x) \,dx Using the property _0^a h(x)\,dx = _0^a h(a-x)\,dx, we have: I = _0^4 f(x) g(x) \,dx Adding the two expressions for I: 2I = _0^4 f(4-x) g(4-x) \,dx + _0^4 f(x) g(x) \,dx Since g(4-x) = g(x), this becomes: 2I = _0^4 f(4-x) + f(x) g(x) \,dx Substituting f(x) + f(4-x) = g(x): 2I = _0^4 g(x) g(x) \,dx = _0^4 1\,dx 2I = [x]_0^4 = 4 I = 2 Answer: 2

Question 4

Passage: Let _0^ 2 a x+b x (a+b)( x+ x) \,dx=k. Question: What is _0^ 2 a x+b x x+ x \,dx equal to ?

Accepted Answer

Given k = _0^ 2 a x+b x (a+b)( x+ x) \,dx Multiplying both sides by (a+b): k(a+b) = _0^ 2 a x+b x x+ x \,dx Using the definite integral property _0^a f(x)\,dx = _0^a f(a-x)\,dx, replacing x with 2 -x: k(a+b) = _0^ 2 a ( 2 -x )+b ( 2 -x ) ( 2 -x )+ ( 2 -x ) \,dx k(a+b) = _0^ 2 a x+b x x+ x \,dx This matches the required integral. Thus, the value is k(a+b). Answer: k(a+b)

Question 5

Consider the following for the two (02) items that follow: Let f(x) = [x^2], where [ ] is the greatest integer function. What is _ 2 ^ 2 f(x) \, dx equal to?

Accepted Answer

The given integral is _ 2 ^ 2 [x^2] \, dx. As x varies from 2 to 2, x^2 varies from 2 to 4. The greatest integer function [x^2] changes its value when x^2 is an integer. The integer values of x^2 in this interval are 2, 3, and 4, which correspond to x = 2 , 3 , and 2. We split the integral at x = 3 : For x ( 2 , 3 ), 2 For x ( 3 , 2), 3 The integral can be written as: _ 2 ^ 2 [x^2] \, dx = _ 2 ^ 3 2 \, dx + _ 3 ^ 2 3 \, dx Evaluating the integrals: _ 2 ^ 3 2 \, dx = 2[x]_ 2 ^ 3 = 2( 3 - 2 ) = 2

Question 6

What is _n^ n+1 (x - [x])\, dx, where [ ] is the greatest integer function and n is a natural number?

Accepted Answer

The given integral is I = _n^ n+1 (x - [x])\, dx. We know that x - [x] = \ x\ , which is the fractional part function. The fractional part function is periodic with a period of 1. Therefore, the integral over any interval of length 1 is equal to the integral over [0, 1]. I = _0^1 (x - [x])\, dx In the interval [0, 1), [x] = 0. I = _0^1 x\, dx = [ x^2 2 ]_0^1 = 1 2 Alternatively, in the interval [n, n+1), [x] = n. I = _n^ n+1 (x - n)\, dx = [ (x-n)^2 2 ]_n^ n+1 = 1^2 2 - 0 = 1 2 Answer: 1 2

Question 7

If I_1 = _e^ e^2 dx x and I_2 = _1^2 e^x x \, dx, then which one of the following is correct?

Accepted Answer

I_1 = _e^ e^2 dx x Substituting x = t x = e^t dx = e^t dt When x = e, t = 1 When x = e^2, t = 2 I_1 = _1^2 e^t t dt Since the definite integral is independent of the variable of integration, I_1 = _1^2 e^x x dx = I_2 I_1 - I_2 = 0 Answer: I_1 - I_2 = 0

Definite Integration — Previous Year Questions & Solutions