Ask a question

Evaluate the integral and interpret it as the area of a region.

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

2 Answers by Expert Tutors

Tutors, sign in to answer this question.
William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
∫[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...
4.9 4.9 (641 lesson ratings) (641)
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