What is #f(x) = int xsinx dx# if #f(pi/3) = -2 #?

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Answer 1 · 2016-11-02T21:01:56

#f(x) = -xcosx + sinx - 2.34#, approximately.

Let #u = x# and #dv = sinxdx#.

So, #du = dx# and #v = -cosx#

Now, use the integration by parts formula #int(udv) = uv - int(vdu)#.

#int(xsinx) = -xcosx - int(-cosx)#

#f(x) = -xcosx + sinx + C#

Now, all we have to do is determine the value of #C#.

We know that when #x = pi/3, y = -2#, so:

#-2 = -pi/3cos(pi/3) + sin(pi/3) + C#

#-2 + pi/3cos(pi/3) + sin(pi/3) = C#

Evaluating using a calculator, you should get #C ~=-2.34#

Hopefully this helps!