# Calculus II

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Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi/2

• Calculus II -

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.

• Calculus II -

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.

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