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>


<math>\frac{d}{dx}\left[h(x)\right]=\frac{-1}{x^2}\left(\arctan(\frac{1}{x}\right)-0</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