`int_-2^2 |x| dx` is equal toA. `4`B. `3.5`C. `2`D. `0`

1.	`int_-2^2 \|x\| dx` is equal toA. `4`B. `3.5`C. `2`D. `0`
Answer» `I=int_(-2)^(0)\|x\|dx+int_(0)^(2)\|x\|dx=int_(-2)^(0)-xdx+int_(0)^(2)xdx`. `=[-(x^(2))/(2)]_(-2)^(0)+[(x^(2))/(2)]_(0)^(2)=0-(-2)+(2-0)=4`.

Discussion

The value of `int_0^oox/((1+x)(x^2+1))dx` isA. `2pi`B. `pi/4`C. `pi/16`D. `pi/32`
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The value of ` int _(0)^(pi//2) ((sin x + cos x)^(2))/(sqrt(1+sin 2x) dx` is
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If `int _(0)^(pi//2) sin^(6) dx = (5pi)/32 ,` then the value of ` int _(-pi)^(pi) (sin ^(6) x + cos^(6)x)dx` isA. `(5pi)/8`B. `(5pi)/16`C. `(5pi)/2`D. `(5pi)/4`
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`l=int_(-2)^(1)(tan^(-1)x+"cot"^(-1)(1)/(x))dx` is equal toA. ` (5pi)/2+ 4 tan ^(-1) 2 - In 5/2`B. ` (5pi)/2 - 4 tan^(-1) 2 + In 5/2`C. ` (5pi)/2 - 3 tan ^(-1) 2 - In 5/2`D. ` (5pi)/2 - 3 tan ^(-1) 2 + 5/2`
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