Integration, help! (1 Viewer)

excellence1 · Feb 13, 2016

Find the area enclosed between the curves y=-x^2 + 2x + 4 and y=x^2.

InteGrand · Feb 13, 2016

excellence1 said:
Find the area enclosed between the curves y=-x^2 + 2x + 4 and y=x^2.

$\noindent Note that the graph of $-x^2 + 2x + 4$ is above that of $y=x^2$ for the region we need. First we find the points of intersection. They occur when and only when$

$\begin{align*}-x^2 + 2x + 4 &= x^2 \\ \Longleftrightarrow -x^2 + 2x + 4-x^2 &= 0 \\ \Longleftrightarrow -2x^2 + 2x + 4 &= 0 \\ \Longleftrightarrow 2x^2 - 2x - 4 &= 0 \\ \Longleftrightarrow x^2 - x - 2 &= 0 \\ \Longleftrightarrow x &= \frac{1\pm \sqrt{1+8}}{2} \\ &= \frac{1\pm 3}{2} \\ &= 2 \text{ or } -1. \end{align*}$

$\noindent So the area $\mathcal{A}$ (in square units) is given by$

$\begin{align*}\mathcal{A} &= \int _{-1} ^{2}\left(-x^2 + 2x + 4-x^2\right)\text{ d}x \\ &= \int _{-1} ^{2}\left(-2x^2 + 2x + 4\right)\text{ d}x \\ &= \left[\frac{-2x^3}{3}+x^2 +4x \right]_{-1}^{2} \\ &= \left(\frac{-2\left(2\right)^3}{3}+(2)^2 +4\times 2 \right)-\left(\frac{-2\left(-1\right)^3}{3}+\left(-1\right)^2 +4\left(-1\right)\right) \\ &= 9.\end{align*}$

Integration, help! (1 Viewer)

excellence1

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InteGrand

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