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>") |
m (Protected "5.4 Indefinite Integrals and the Net Change Theorem/27" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))) |
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<math>\ | <math> | ||
\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{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{5}{2}}}{5}\right)\Bigg|_{1}^{4} \\[2ex] | ||
= | &=\left[\frac{2(4)^{\frac{3}{2}}}{3}+\frac{2(4)^{\frac{5}{2}}}{5}\right]-\left[\frac{2(1)^{\frac{3}{2}}}{3}+\frac{2(1)^{\frac{5}{2}}}{5}\right] \\[2ex] | ||
= | &=\left[\frac{16}{3}+\frac{64}{5}\right]-\left[\frac{2}{3}+\frac{2}{5} \right] \\[2ex] | ||
= | &=\frac{256}{15} | ||
\end{align} | |||
</math> | |||
Latest revision as of 19:40, 21 September 2022
![{\displaystyle {\begin{aligned}\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 {2t^{\frac {3}{2}}}{3}}+{\frac {2t^{\frac {5}{2}}}{5}}\right){\Bigg |}_{1}^{4}\\[2ex]&=\left[{\frac {2(4)^{\frac {3}{2}}}{3}}+{\frac {2(4)^{\frac {5}{2}}}{5}}\right]-\left[{\frac {2(1)^{\frac {3}{2}}}{3}}+{\frac {2(1)^{\frac {5}{2}}}{5}}\right]\\[2ex]&=\left[{\frac {16}{3}}+{\frac {64}{5}}\right]-\left[{\frac {2}{3}}+{\frac {2}{5}}\right]\\[2ex]&={\frac {256}{15}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/fda02451a09d311d14ed3b2910293328b5ffe18a)