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

Question

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

Scarlett Allison · Accepted Answer

#f(x)=-cosx-x^2sinx-2xcosx+2sinx+1/72(72-36sqrt3-49pi^2-84sqrt3pi)#.

#f(x)=intsinxdx-intx^2cosxdx=-cosx-I,# where, #I=intx^2cosxdx=intuvdx", say,"# where, #u=x^2, and, v=cosx#. But, by the Rule of Integration by Parts (ibp), #intuvdx=uintvdx-int((du)/dxintvdx)dx# Here, #intvdx=intcosxdx=sinx, and, (du)/dx=2x#. #:. I=x^2sinx-int(2xsinx)dx# #=x^2sinx-2J", where, "J=intxsinxdx#. For, #J#, we again use ibp; this time, with #u=x, &, v=sinx#. #:. J=x(intsinxdx)-int{d/dx(x)*intsinxdx}dx# #=x(-cosx)-int{(1)(-cosx)}dx#, i.e., #J=-xcosx+intcosxdx=-xcosx+sinx# Thus, altogether, we have, #I=x^2sinx-2J=x^2sinx-2{-xcosx+sinx}#, or, #I=x^2sinx+2xcosx-2sinx#, so that, finally, #f(x)=-cosx-I,# #=-cosx-{x^2sinx+2xcosx-2sinx}#, i.e., #f(x)=-cosx-x^2sinx-2xcosx+2sinx+C#. To determine #C#, we use the cond. : #f(7pi/6)=0# #rArr -cos(7pi/6)-49pi^2/36sin(7pi/6)-7pi/3cos(7pi/6)+2sin(7pi/6)+C=0#. #rArrsqrt3/2+49pi^2/36*1/2+7pi/3*sqrt3/2-1+C=0# #rArr36sqrt3+49pi^2+84sqrt3pi-72+72C=0# #rArr C=1/72(72-36sqrt3-49pi^2-84sqrt3pi)#. Therefore, #f(x)=-cosx-x^2sinx-2xcosx+2sinx+1/72(72-36sqrt3-49pi^2-84sqrt3pi)#. Enjoy Maths.!

Evelyn Agard · Answer

undefined To find the value of $ f(x) = \int \sin(x) - x^2 \cos(x) \, dx $ when $ f\left(\frac{7\pi}{6}\right) = 0 $, you first need to find the antiderivative of $ \sin(x) - x^2 \cos(x) $, then evaluate it at $ x = \frac{7\pi}{6} $. The antiderivative is $ F(x) = -\cos(x) - \frac{x^3}{3} $. Then, plug in $ x = \frac{7\pi}{6} $ into $ F(x) $ and solve for $ f\left(\frac{7\pi}{6}\right) $.