Thursday

November 20, 2014

November 20, 2014

Posted by **Ethan** on Wednesday, September 28, 2011 at 12:00am.

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 -
**Steve**, Wednesday, September 28, 2011 at 10:46amThis 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

**Answer this Question**

**Related Questions**

Please Help!!! - given the graph of f(x) = x sinx, 0<=x<=2pi assuming ...

calculus - given the graph of f(x) = x sinx, 0<=x<=2pi assuming that a ...

Math, Calculus - Find the derivative of the function. y= xcosx - sinx What's the...

math - Hello there, im doing an equation and its mind boggling me badly so i ...

Calculus - Find the area between y=cosx and y=sinx from 0 to 2pi. To find the ...

math - the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity ...

calculus - 3. Let f be the function defined by f(x)=ln(2+sinx) for pi<=x<=...

calculus - find all extrema in the interval [0, 2pi] if y=sinx + cosx and also a...

college/microeconomics - Just needing to know if I have done the work correctly ...

Microeconomics: - Just needing to know if I have done the work correctly with ...