# please show how to calculate definite integral of y=|sin x-square root 3*cos x|, in interval [0,pi/2]?

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### 1 Answer

The absolute value forces the answers to definite integrals to be positive on the interval [0,pi/2]. Therefore the integral needs to be split up. Looking at the graph of sin x - sqrt 3 (cos x), from [0,pi/3] the y-values are negative. From [pi/3, pi/2] the y-values are positive.

The definite integral breaks into two.

`-int sin(x)-sqrt(3)cos(x)dx` from [0,pi/3] plus

`int sin(x)-sqrt(3)cos(x)dx` from [pi/3, pi/2]

The integral is equal to -cos(x)-sqrt(3)*sin(x)

Putting in the bounds:

-[-cos(pi/3)-sqrt(3)sin(pi/3)-(-cos(0)-sqrt(3)sin(0))]+

[-cos(pi/2)-sqrt(3)sin(pi/2)-(-cos(pi/3)-sqrt(3)sin(pi/3))] =

-[-1/2-3/2-(-1-0)] + [-0-sqrt(3)-(-1/2-3/2)] =

-[-1]+[2-sqrt(3)] = 3-sqrt(3)

The exact answer is 3-sqrt(3) which is approximately 1.268.