5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions
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\int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex] | \int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex] | ||
&=\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5} | &=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)|_{1}^{4} \\[2ex] | ||
=\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4} | =\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4} | ||
Revision as of 15:08, 21 September 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex] &=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)|_{1}^{4} \\[2ex] =\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4} =\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5} =\frac{256}{15} \end{align} }