# calculus

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given the graph of f(x) = x sinx, 0<=x<=2pi
assuming that a quantity y changes at a rate of y' = xsinx, find by how much it will increase or decrease over 3pi/2 <= x <= 2pi

• calculus - ,

This is just a problem in integrals. We have y', and we want to add up all the small changes given by that function.

Int[x sinx dx] can be solved using integration by parts.

Recall the product rule of derivatives:

(uv)' = u'v + uv'
so
uv' = (uv)' - u'v

Going the other direction,

Int[u dv] = uv - Int[v du]

So, we want to split up the integrand into two factors, where one part (u) gets simpler after differentiation, and the other part (dv) can be easily integrated.

Here, we have x sinx
If u = x, then du = dx
If dv = sinx, then v = -cosx

Int[x sinx dx] = uv - Int[v du]
= -x cosx - Int[-cos x * dx]

Now we have a simple integrand, -cosx

Int cos x dx = sin x
So, the final integration is

Int[x sinx dx] = uv - Int[v du]
= -x cosx - Int[-cos x * dx]
= -x cosx + sin x

Evaluating at 3pi/2 and 2pi, we have

(-2pi * 1 + 0) - (-3pi/2 * 0 + -1) = -2pi + 3pi/2 + 1 = 1 + 7pi/2