5.3 The Fundamental Theorem of Calculus/33: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| (57 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
< | <math> | ||
\begin{align} | |||
\int_{1}^{2}\left(1+2y\right)^2dy &=\int_{1}^{2}\left(1+4y+4y^2\right)dy \\[2ex] | |||
&=y+\frac{4y^2}{2}+\frac{4y^3}{3}\bigg|_{1}^{2} \\[2ex] | |||
&=\left[2+\frac{4(2)^2}{2}+\frac{4(2)^3}{2}\right]-\left[1+\frac{4(1)^2}{2}+\frac{4(1)^3}{2}\right] \\[2ex] | |||
&=\frac{49}{3} | |||
\end{align} | |||
</math> | |||
Latest revision as of 21:32, 6 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{2}\left(1+2y\right)^{2}dy&=\int _{1}^{2}\left(1+4y+4y^{2}\right)dy\\[2ex]&=y+{\frac {4y^{2}}{2}}+{\frac {4y^{3}}{3}}{\bigg |}_{1}^{2}\\[2ex]&=\left[2+{\frac {4(2)^{2}}{2}}+{\frac {4(2)^{3}}{2}}\right]-\left[1+{\frac {4(1)^{2}}{2}}+{\frac {4(1)^{3}}{2}}\right]\\[2ex]&={\frac {49}{3}}\end{aligned}}}