Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 48

Find the definite integral $\displaystyle \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx$

Let $u = 1 + 2x$, then $du = -2 dx$, so $\displaystyle dx = \frac{du}{2}$. When $x = 0, u =1$ and when $x = 4, u = 9$. Thus,



$\begin{equation} \begin{aligned} \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \int^4_0 \frac{\displaystyle \frac{u - 1}{2}}{\sqrt{u}} \frac{du}{2} \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \int^4_0 \frac{u - 1}{2 \sqrt{u}} \cdot \frac{du}{2} \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} \int^4_0 \frac{u - 1}{\sqrt{u}} du \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} \int^4_0 \frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}} du \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} \int^4_0 u^{\frac{1}{2}} - u^{\frac{-1}{2}} du \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} \left[ \frac{u^{ \frac{1}{2}+ 1}}{\displaystyle \frac{1}{2} + 1} - \frac{u^{ \frac{-1}{2}+ 1}}{\displaystyle \frac{-1}{2} + 1} \right]^4_0 \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} \left[ \frac{u^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} - \frac{u^{\frac{-1}{2}}}{\displaystyle \frac{1}{2}} \right]^4_0 \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} \left[ \frac{2(9)^{\frac{3}{2}}}{3} - 2 (9)^{\frac{1}{2}} \right] - \frac{1}{4} \left[ \frac{2(1)^{\frac{3}{2}}}{3} - 2(1)^{\frac{1}{2}} \right] \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{1}{4} (12) - \frac{1}{4} \left(- \frac{4}{3} \right) \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& 3 + \frac{4}{12} \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& 3 + \frac{1}{3} \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{9 + 1}{3} \\ \\ \int^4_0 \frac{x}{\sqrt{1 + 2x}} dx =& \frac{10}{3} \end{aligned} \end{equation}$