The value of the integral ` int _(0)^(2) | x^(2)-1| dx` is

1.	The value of the integral ` int _(0)^(2) \| x^(2)-1\| dx` is
Answer» Correct Answer - B ` int _(0)^(2) \| x^(2) -1\| dx = int _(0)^(1) - (x^(2) -1) dx + int _(1)^(2) (x^(2) -1)dx ` ` = - int _(0)^(1) (x^(2) -1) dx + int _(1)^(2) (x^(2) -1) dx = ((x^(3))/3-x)_(0)^(1) + ((x^(3))/3-x)^(2)` , ` = ((x^(3))/3 - x)_(0)^(1) + ((x^(3))/3 - x)_(1)^(2) = 2 - 1/3 + 1 + 8/3 -2 -1/3 + 1 = 2`

Discussion

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The value of ` int _(0)^(pi//2) ((sin x + cos x)^(2))/(sqrt(1+sin 2x) dx` is
` int _(-3pi//2)^(-pi//2) [ ( x + pi)^(3) + cos^(2) x ] ` dx is equalt toA. `((pi^(4))/32)+(pi/2)`B. ` (pi/2)`C. ` (pi/4)-1`D. ` (pi^(4))/32`
The value of ` int _(0)^(pi) (dx)/(5+4 cos x ) ` isA. `2pi`B. `(3pi)/2`C. `(5pi)/4`D. `pi/3`
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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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