What is $f (x) = \int x sin x d x$ $f(x) = int xsinx dx$ if $f (\frac{π}{3}) = - 2$ $f(pi/3) = -2$ ?

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Answer 1 · 2016-11-02T21:01:56

$f (x) = - x cos x + sin x - 2.34$ $f(x) = -xcosx + sinx - 2.34$ , approximately.

Let $u = x$ $u = x$ and $d v = sin x d x$ $dv = sinxdx$ .

So, $d u = d x$ $du = dx$ and $v = - cos x$ $v = -cosx$

Now, use the integration by parts formula $\int (u d v) = u v - \int (v d u)$ $int(udv) = uv - int(vdu)$ .

$\int (x sin x) = - x cos x - \int (- cos x)$ $int(xsinx) = -xcosx - int(-cosx)$

$f (x) = - x cos x + sin x + C$ $f(x) = -xcosx + sinx + C$

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

We know that when $x = \frac{π}{3}, y = - 2$ $x = pi/3, y = -2$ , so:

$- 2 = - \frac{π}{3} cos (\frac{π}{3}) + sin (\frac{π}{3}) + C$ $-2 = -pi/3cos(pi/3) + sin(pi/3) + C$

$- 2 + \frac{π}{3} cos (\frac{π}{3}) + sin (\frac{π}{3}) = C$ $-2 + pi/3cos(pi/3) + sin(pi/3) = C$

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

Hopefully this helps!

What is f(x) = int xsinx dxf(x)=∫xsinxdxf(x) = int xsinx dx if f(pi/3) = -2 f(π3)=−2f(pi/3) = -2 ?