Then $dx = \sqrt{24} \sec^2 \theta d\theta$.
When $x = 1$, $\tan \theta = \frac{1}{\sqrt{24}}$, so $\theta = \arctan(\frac{1}{\sqrt{24}})$.
When $x = b$, $\tan \theta = \frac{b}{\sqrt{24}}$, so $\theta = \arctan(\frac{b}{\sqrt{24}})$.
Then
\begin{align*}
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