diff --git a/College/AU 26/MATH 224 (Multivar)/Class 10-5.md b/College/AU 26/MATH 224 (Multivar)/Class 10-5.md
index 3dc4fd1..87e6348 100644
--- a/College/AU 26/MATH 224 (Multivar)/Class 10-5.md	
+++ b/College/AU 26/MATH 224 (Multivar)/Class 10-5.md	
@@ -15,4 +15,5 @@ $$\begin{gathered} 1<x<2 \\[1.5ex] x < y < x^2 \end{gathered}$$
 We can think of finding the volume of the solid by using slices parallel to the $YZ$ plane. With the domain provided this area can be found with 
 $$ Area(x^*)=\int_{x}^{x^2}x^*ydy$$
 Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as 
-$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \int_{x}^{x^2} \frac{xy^2}{2}   \right] dx \end{gather}$$
+$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \  dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \  \right] dx \\[1.5ex]  \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2)  \right] dx \\[1.5ex]  \int_{1}^2  \frac{x^5}{2}-\frac{x^3}{2}  dx \\[1.5ex]   \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right)  \end{gather}$$
+27/8
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