What is #f(x) = int 3sinx-xcosx dx# if #f((7pi)/8) = 0 #?

Question

What is #f(x) = int 3sinx-xcosx   dx# if #f((7pi)/8) = 0  #?

Scarlett Allison · Accepted Answer

#f(x) = -cosx - xsinx - 3cosx - 2.64#, nearly.

Separate the integrals. #f(x) = int3sinxdx - intxcosxdx# Integrate #intxcosxdx# by parts. Let #u = x# and #dv= cosxdx#. Then #du = dx# and #v= sinx#. #intudv = uv - intvdu# #intxcosx = xsinx - intsinxdx# #intxcosx = xsinx - (-cosx) + C# #intxcosx = xsinx + cosx + C# Put this together: #f(x) = int3sinxdx - (xsinx + cosx) + C# #f(x) = int3sinxdx - xsinx - cosx + C# #f(x) = C - 3cosx - xsinx - cosx # You can solve for #C# now. We know that when #x= (7pi)/8#, #y = 0#. #0 = C - 3cos((7pi)/8) - (7pi)/8sin((7pi)/8) - cos((7pi)/8)# #C = 3cos((7pi)/8) + (7pi)/8sin((7pi)/8) + cos((7pi)/8)# This will not be an exact expression. An approximation using a calculator yields #C ~~ -2.64#. Therefore, #f(x) = -cosx - xsinx - 3cosx - 2.64#. Hopefully this helps!

Cameron Alexander · Answer

undefined To find the value of $ f(x) $ when $ f\left(\frac{7\pi}{8}\right) = 0 $, we first need to evaluate the definite integral of $ 3\sin(x) - x\cos(x) $ from some initial value to $ \frac{7\pi}{8} $ and set it equal to 0.

Given:
$$ f(x) = \int (3\sin(x) - x\cos(x)) \, dx $$

We need to find $ f(x) $ when $ x = \frac{7\pi}{8} $ and $ f\left(\frac{7\pi}{8}\right) = 0 $. This means the integral from some initial value to $ \frac{7\pi}{8} $ equals 0.

Solving the integral and finding the antiderivative, we get:
$$ f(x) = -3\cos(x) - x\sin(x) + C $$

Now, plug in $ x = \frac{7\pi}{8} $ and set it equal to 0 to find $ C $:
$$ -3\cos\left(\frac{7\pi}{8}\right) - \frac{7\pi}{8}\sin\left(\frac{7\pi}{8}\right) + C = 0 $$

Now, solve for $ C $:
$$ C = 3\cos\left(\frac{7\pi}{8}\right) + \frac{7\pi}{8}\sin\left(\frac{7\pi}{8}\right) $$

Therefore, the function $ f(x) $ is:
$$ f(x) = -3\cos(x) - x\sin(x) + 3\cos\left(\frac{7\pi}{8}\right) + \frac{7\pi}{8}\sin\left(\frac{7\pi}{8}\right) $$