Find Vol Bound by $x_2 =\frac{y+1}{2}$ & $x_1=y^2$

[karush]

Find volume bound by $x_2 =\frac{y+1} {2} $ and $x_1=y^2$
$\int_{-{\frac{1}{2}}}^{1} (-{y}^{2 }+\frac{y}{2}+\frac{1}{2}) \,dx$
$
\begin{array}{l}
{{\left[{\frac{{y}^{2}}{2}\mathrm{{+}}\frac{y}{4}\mathrm{{-}}\frac{{y}^{3}}{3}}\right]}_{\mathrm{{-}}{1}{\mathrm{/}}{2}}^{1}}\\
{\left[{\frac{\frac{1}{4}}{2}\mathrm{{-}}\frac{\frac{1}{2}}{4}\mathrm{{+}}\frac{\frac{1}{8}}{3}}\right]\mathrm{{-}}\left[{\frac{1}{4}\mathrm{{+}}\frac{1}{2}\mathrm{{-}}\frac{1}{3}}\right]}\\
{\left[{\frac{3}{\mathrm{48}}\mathrm{{-}}\frac{\mathrm{12}}{\mathrm{48}}\mathrm{{+}}\frac{2}{\mathrm{48}}}\right]\mathrm{{-}}\left[{\frac{\mathrm{12}}{\mathrm{48}}\mathrm{{+}}\frac{\mathrm{24}}
{\mathrm{48}}\mathrm{{-}}\frac{\mathrm{16}}{\mathrm{48}}}\right]}
\end{array}$
$$
\frac{5}{12 }+\frac{7}{48 }=\frac{9}{16}
$$
Thot I would try mathmajic but better to stay here

 

[Prove It]

Find volume bound by $x_2 =\frac{y+1} {2} $ and $x_1=y^2$
$\int_{-{\frac{1}{2}}}^{1} (-{y}^{2 }+\frac{y}{2}+\frac{1}{2}) \,dx$
$
\begin{array}{l}
{{\left[{\frac{{y}^{2}}{2}\mathrm{{+}}\frac{y}{4}\mathrm{{-}}\frac{{y}^{3}}{3}}\right]}_{\mathrm{{-}}{1}{\mathrm{/}}{2}}^{1}}\\
{\left[{\frac{\frac{1}{4}}{2}\mathrm{{-}}\frac{\frac{1}{2}}{4}\mathrm{{+}}\frac{\frac{1}{8}}{3}}\right]\mathrm{{-}}\left[{\frac{1}{4}\mathrm{{+}}\frac{1}{2}\mathrm{{-}}\frac{1}{3}}\right]}\\
{\left[{\frac{3}{\mathrm{48}}\mathrm{{-}}\frac{\mathrm{12}}{\mathrm{48}}\mathrm{{+}}\frac{2}{\mathrm{48}}}\right]\mathrm{{-}}\left[{\frac{\mathrm{12}}{\mathrm{48}}\mathrm{{+}}\frac{\mathrm{24}}
{\mathrm{48}}\mathrm{{-}}\frac{\mathrm{16}}{\mathrm{48}}}\right]}
\end{array}$
$$
\frac{5}{12 }+\frac{7}{48 }=\frac{9}{16}
$$
Thot I would try mathmajic but better to stay here

9/16 is correct. Well done.

 

[karush]

It's the arithmetic that kills on these integrals