`f(x) = {(1-2x", for " x le0),(1+2x", for "xgt0):}` then `int_(-1) ^(1) f(x) dx=`A. 4B. `-4`C. 2D. `-2`

`f(x) = {(1-2x", for " x le0),(1+2x", for "xgt0):}` then `int_(-1) ^(1

1.	`f(x) = {(1-2x", for " x le0),(1+2x", for "xgt0):}` then `int_(-1) ^(1) f(x) dx=`A. 4B. `-4`C. 2D. `-2`
Answer» Correct Answer - A

`int_(0)^((pi)/(4))(dx)/(acos^(2)x+bsin^(2)x)=`A. `(1)/(ab)tan^(-1)((b)/(a))" if "agt0,bgt0`B. `(1)/(ab)tan^(-1)((b)/(a))" if "alt0,blt0`C. `(pi)/(4)" if "a=1, b=1`D. none of these
The value of `int_(0)^((pi)/(4))tan^(2)thetad theta` is-A. `(pi)/(4)-1`B. `1-(pi)/(4)`C. `(pi)/(2)-1`D. `pi-2`
`int_0^(pi/4) (sinx+cosx)/(9+16sin2x)dx`A. `1/10 log 3`B. `1/20log3`C. `1/10log5`D. `1/20log5`
integrate `int_0^(2pi) e^x . sin (pi/4 + x/2) dx`A. `(-3sqrt(2))/ 5 (e^(2pi)-1)`B. `(3sqrt(2))/ 5 (e^(2pi)-1)`C. `(-3sqrt(2))/ 5 (e^(2pi)+1)`D. `(3sqrt(2))/ 5 (e^(2pi)+1)`
`int_0^1(cos^(- 1)x)^2dx`A. `pi/2 +1`B. `pi/2-1`C. `pi+2`D. `pi-2`
`int_0^pisin^3x(1+2cosx)(1+cosx)^2dx`A. `8/3`B. `(-8)/3`C. `4/3`D. `(-4)/3`
If `int_0^a3x^2dx=8,`find the value of `adot`
The integral `int_0^pisqrt(1+4sin^2x/2-4sinx/2dx)`equal(1) `pi-4`(2) `(2pi)/3-4-4sqrt(3)`(3) `4""sqrt(3)-4`(4) `4""sqrt(3)-4-pi/3`A. `pi-4`B. `4sqrt(3)-4`C. `(2pi)/3 -4-4sqrt(3)`D. `4sqrt(3) - 4 - pi/3`
Evaluate :`int_0^(pi/2)(dx)/(1+sqrt(tanx))`A. `(3pi)/2`B. `(3pi)/4`C. `pi/2`D. `pi/4`
`int_0^(pi//2)(theta/sintheta)^2d theta=`A. `pi/(log2)`B. `log 2`C. `pi log 2`D. `pi`