Calculus II
posted by Scott on .
Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi/2

well, the first question is do they cross?
xsinx=x
sinx=1
x=PI/2
so no crossing.
Then area is
int (xxsinx)dx over limits.
x^2/2 sinx+xcosx
PI/2)^2/21 check that. 
I so happens that the two functions intersect at
x = 0 and x = π/2, the domain of our area
area
= [integral] (xsinx  x)dx from 0 to π/2
= sinx  xcosx  (1/2)x^2  from 0 to π/2
= sinπ/2  π/2(cosπ/2)  π^2/8  (sin0  0  0)
= 1  0  π^2/8
= 1  π^2/8
=  .2337
OOPS, just realized that my assumption that the trig curve was above the straight line was false, so we have to reverse the integrand
area = integral (x  xsinx)dx
make the necessary changes, only the signs will be affected, or
we could just take the absolute value of each of my lines.