I can't figure out where I'm going wrong with the following integral:
Given: $$ V = \pi\int_1^2 (2-\frac{x}{2})^2 dx$$
Substitutions: $$ u = 2-\frac{x}{2} $$ $$ du = -\frac{1}{2} dx$$ $$ (-2)du = dx $$
Evaluate new limits of integration: $$ 2-\frac{1}{2}(1) = \frac{3}{2} $$ $$ 2-\frac{1}{2}(2) = 1 $$
Integrating in terms of u: $$ \therefore \pi\int_1^2 (2-\frac{x}{2})^2 dx = \pi\int_\frac{3}{2}^1 u^2 (-2)du$$ $$ = -2\pi\left.(\frac{u^3}{3}\right|_\frac{3}{2}^1 )$$ $$ = -2\pi(\frac{1}{3} - \frac{27}{8}) = -2\pi(\frac{8-81}{24}) = -2\pi(\frac{-73}{24})$$ $$ = \frac{73\pi}{12} $$
The correct answer, which I was able to get when I don't use u-substitution, is: $$ \frac{19\pi}{12} $$
Can someone please point out where I went wrong?
