5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions

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(Created page with "<math>\int_{1}^{4}\sqrt{t}(1+t)dt</math> =<math>\int_{}^{}\sqrt{t}(1+t)dt</math> =<math>\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}</math> =<math>\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4}</math> =<math>\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}</math> =<math>\frac{256}{15}</math>")
 
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<math>\int_{1}^{4}\sqrt{t}(1+t)dt</math>
<math>
\begin{align}


=<math>\int_{}^{}\sqrt{t}(1+t)dt</math>
\int_{1}^{4}\sqrt{t}(1+t)dt =\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt


=<math>\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}</math>
=\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}


=<math>\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4}</math>
=\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4}


=<math>\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}</math>
=\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}


=<math>\frac{256}{15}</math>
=\frac{256}{15}
 
\end{align}
</math>

Revision as of 15:07, 21 September 2022