Definite integrals by substitution
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Evaluate \(\int\limits_0^3 {t\sqrt {t + 1} } \,dt\)
\begin{aligned}u=t+1 \quad & t=3 \rightarrow u=4 \\\frac{d u}{d t}=1\quad & t=0 \rightarrow u=1 \\du=dt\quad &\\\int_{0}^{3} t \sqrt{t+1} d t&= \int_{1}^{4}(u-1) \sqrt{u} \,d u \\& =\int_{1}^{4} u^{\frac{3}{2}}-u^{\frac{1}{2}} d u \\& =\left[\frac{2}{5} u^{\frac{5}{2}}-\frac{2}{3} u^{\frac{3}{2}}\right]_{1}^{4} \\& =\left[\left(\frac{2}{5} \times 4^{\frac{5 } {2}}-\frac{2}{3} 4^{\frac{3}{ 2}}\right)-\left(\frac{2}{5}-\frac{2}{3}\right)\right] \\& =\left[\frac{64}{5}-\frac{16}{3}-\frac{2}{5}+\frac{2}{3}\right] \\& =7 \frac{11}{15}\end{aligned}
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