well, the first question is do they cross?
xsinx=x
sinx=1
x=PI/2
so no crossing.
Then area is
int (x-xsinx)dx over limits.
x^2/2 -sinx+xcosx
PI/2)^2/2-1 check that.
Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi/2
2 answers
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.
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.
1 answer