What is #f(x) = int 2sinx-xcosx dx# if #f((7pi)/6) = 0 #?

Question

What is #f(x) = int 2sinx-xcosx   dx# if #f((7pi)/6) = 0  #?

Ezekiel Alexander · Accepted Answer

#f(x)=-3cosx-xsinx-(3sqrt3)/2-(7pi)/12# Split up the integral into #f(x)=2intsinxdx-intxcosxdx# Using integration by parts, we get #f(x)=-2cosx-xsinx-cosx+C# #f(x)=-3cosx-xsinx+C# Substituting #x=(7pi)/6#, we get #f((7pi)/6)=0=-3cos((7pi)/6)-(7pi)/6sin((7pi)/6)+C# Solving for C, we get #C=-(3sqrt3)/2-(7pi)/12# Therefore, #f(x)=-3cosx-xsinx-(3sqrt3)/2-(7pi)/12#.

Cameron Alexander · Answer

undefined The function $ f(x) = \int (2\sin(x) - x\cos(x)) \, dx $ satisfies $ f\left(\frac{7\pi}{6}\right) = 0 $.

Paisley Johnson · Answer

undefined To find $ f(x) = \int (2 \sin(x) - x \cos(x)) \, dx $ when $ f\left(\frac{7\pi}{6}\right) = 0 $, first, find the antiderivative of $ 2\sin(x) - x\cos(x) $. Then use the given value to solve for the constant of integration. Finally, evaluate the definite integral at $ \frac{7\pi}{6} $.

The antiderivative of $ 2\sin(x) - x\cos(x) $ is $ -2\cos(x) - x\sin(x) + C $, where $ C $ is the constant of integration.

Since $ f\left(\frac{7\pi}{6}\right) = 0 $, we have:
$$ 0 = -2\cos\left(\frac{7\pi}{6}\right) - \frac{7\pi}{6}\sin\left(\frac{7\pi}{6}\right) + C $$

Now, evaluate the expression and solve for $ C $:
$$ 0 = -2\left(-\frac{\sqrt{3}}{2}\right) - \frac{7\pi}{6}\left(-\frac{1}{2}\right) + C $$
$$ 0 = \sqrt{3} + \frac{7\pi}{12} + C $$
$$ C = -\sqrt{3} - \frac{7\pi}{12} $$

Substitute $ C $ back into the antiderivative:
$$ f(x) = -2\cos(x) - x\sin(x) - \sqrt{3} - \frac{7\pi}{12} $$

Evaluate the definite integral at $ x = \frac{7\pi}{6} $:
$$ f\left(\frac{7\pi}{6}\right) = -2\cos\left(\frac{7\pi}{6}\right) - \frac{7\pi}{6}\sin\left(\frac{7\pi}{6}\right) - \sqrt{3} - \frac{7\pi}{12} $$

Since $ f\left(\frac{7\pi}{6}\right) = 0 $, we have:
$$ 0 = -2\left(-\frac{\sqrt{3}}{2}\right) - \frac{7\pi}{6}\left(-\frac{1}{2}\right) - \sqrt{3} - \frac{7\pi}{12} $$
$$ 0 = \sqrt{3} + \frac{7\pi}{12} - \sqrt{3} - \frac{7\pi}{12} $$

Hence, the value of $ f\left(\frac{7\pi}{6}\right) $ is indeed 0.