an integral from 0 to pi/2 the function is |5sin(x)-5cos(2x)|dx

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∫[5sin(x) - 5cos(2x)]dx = (-5/2)sin(2x) - 5 cos(x) + c

Evaluated between 0 and (pi/2) this becomes 5.

This represents the area of the curve between 0 and (pi/2).

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...

This integral is a bit complicated because of the absolute value. The argument of the abs function is negative (or zero) in the range 0 to π/6 and positive (or zero) in the range π/6 to π/2. We must break up the range of integration into two parts: 0 to π/6 and then π/6 to π/2.

The integrand in the first part is 5(cos(2x) - sin(x) ) and the integrand in the second part is

5 ( sin(x) - cos(2x) ). We must evaluate the two definite integrals and then add to get the final answer.

The antiderivative of sin(x) is - cos(x) and the antiderivative of cos(2x) is sin(2x) /2.

When all the pieces are put together, the final result is 5 ( (3/2) sqrt(3) -1 ) ~ 7.99

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