5.3 The Fundamental Theorem of Calculus/33: Difference between revisions

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<math>h(x)=\int_{2}^{1/x}\arctan(t)dt</math>
<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 (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}^{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} }