Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 36

Find the definite integral $\displaystyle \int^7_0 \sqrt{4 + 3x} dx$

Let $u = 4 + 3x$, then $du = 3 dx$, so $\displaystyle dx = \frac{du}{3}$. When $x = 0, u = 4$ and when $x = 7, u = 25$. Thus,


$
\begin{equation}
\begin{aligned}

\int^7_0 \sqrt{4 + 3 x} dx =& \int^7_0 \sqrt{u} \frac{du}{3}
\\
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{1}{3} \int^7_0 u^{\frac{1}{2}} du
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{1}{3} \left[ \frac{u^{ \frac{1}{2}+1}}{\displaystyle \frac{1}{2} + 1} \right]^7_0
\\
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{1}{3} \left[ \frac{u^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} \right]^7_0
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{1}{3} \left[ \frac{2u^{\frac{3}{2}}}{3} \right]^7_0
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{2u^{\frac{3}{2}}}{9} \left|^7_0 \right.
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{2 (25)^{\frac{3}{2}}}{9} - \frac{2(4)^{\frac{3}{2}}}{9}
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\int^7_0 \sqrt{4 + 3 x} dx =&\frac{250 - 16}{9}
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\int^7_0 \sqrt{4 + 3 x} dx =& \frac{234}{9}
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\int^7_0 \sqrt{4 + 3 x} dx =& 26
\end{aligned}
\end{equation}
$