If $\displaystyle \int \limits_{0}^{a} \sqrt{x} \; dx = 2 a \int \limits_{0}^{\frac{\pi}{2}} \sin^3 x \; dx$, then obtain $\displaystyle \int \limits_{a}^{a + 1} x \; dx$
$\displaystyle \int \limits_{0}^{a} \; \sqrt{x} \; dx = \dfrac{2}{3} \left[x^{3/2}\right]_{0}^{a} = \dfrac{2}{3} a^{3/2}$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
\int \limits_{0}^{\frac{\pi}{2}} \sin^3 x \; dx & = \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{3 \sin x - \sin 3x}{4}\right) \; dx \hspace{3em} \left[\text{Note: }\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta \right] \\\\
& = \dfrac{3}{4} \int \limits_{0}^{\frac{\pi}{2}} \sin x \; dx - \dfrac{1}{4} \int \limits_{0}^{\frac{\pi}{2}} \sin 3x \; dx \\\\
& = \dfrac{3}{4} \left[- \cos x\right]_{0}^{\pi / 2} - \dfrac{1}{4} \left[\dfrac{- \cos 3x}{3}\right]_{0}^{\pi / 2} \\\\
& = - \dfrac{3}{4} \left(0 - 1\right) + \dfrac{1}{12} \left(0 - 1\right) \\\\
& = \dfrac{3}{4} - \dfrac{1}{12} = \dfrac{2}{3} \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore$ $\;$ As per the question, we have from equations $(1)$ and $(2)$,
$\dfrac{2}{3} a^{3/2} = 2 a \times \dfrac{2}{3} = \dfrac{4a}{3}$
i.e. $a^{3/2 - 1} = 2$
i.e. $a^{1/2} = 2$ $\implies$ $a = 4$ $\;\;\; \cdots \; (3)$
$\therefore$ $\;$ In view of equation $(3)$ we have
$\displaystyle \int \limits_{a}^{a+1} x \; dx = \int \limits_{4}^{5} x \; dx = \left[\dfrac{x^2}{2}\right]_{4}^{5} = \dfrac{1}{2} \left(25 - 16\right) = \dfrac{9}{2}$