`int_-2^2 |1-x^2|dx=` (A) `4` (B) `2` (C) `-2` (D) `0`

AI Generated Solution

Similar Questions

Explore conceptually related problems

The value of the integral int_(-2)^2|1-x^2|dx is 4 b. 2 c. -2 d. 0

int_-1^1 sin^-1(x/(1+x^2))dx= (A) pi/4 (B) pi/2 (C) pi (D) 0

int_0^oo (xlogx)/(1+x^2)^2dx= (A) e (B) 1 (C) 0 (D) none of these

int_0^pi sqrt((1+cos2x)/2)dx= (A) 0 (B) 1 (C) 2 (D) none of these

Let f(x)=max. {2-x,2,1+x} then int_(-1)^1 f(x)dx= (A) 0 (B) 2 (C) 9/2 (D) none of these

The value of int_(-2)^(3)(|x|)/(x)dx is (a) 5 (b) 2 (c) -2 (d) 3

int_(log(1/2))^(log2) log(x+sqrt(x^2+1))dx= (A) 2log2 (B) 1-log2 (C) 1+log2 (D) 0

The value of the integral int_-a^a (x e^(x^2))/(1+x^2)dx is (A) e^(a^2) (B) 0 (C) e^(-a^2) (D) a

[ If f(x)=|x|+|x-1|, then int_(0)^(2)f(x)dx equals- [ (A) 3, (B) 2, (C) 0, (D) -1]]

The value of int_-1^3 [tan^-1(x/(x^2+1))+tan^-1((x^2+1)/x)]dx= (A) pi/2 (B) 2pi (C) pi (D) pi/4