Need help with integration (1 Viewer)

excellence1 · Feb 7, 2016

I did this question like 10 times, i dont know what im doing wrong. The question is

Find the area enclosed between the curves y= (x-1)^2 and y= 5-x^2

InteGrand · Feb 7, 2016

excellence1 said:
I did this question like 10 times, i dont know what im doing wrong. The question is

Find the area enclosed between the curves y= (x-1)^2 and y= 5-x^2

$\noindent Observe from a sketch that the graph of $y= 5-x^2$ is the above that of $y=(x-1)^2$ for the region whose area we wish to calculate. We need to find the $x$-values of the points of intersection of the two curves. To do this, let $(x-1)^2 = 5-x^2$, so$

$\begin{align*}x^2 - 2x + 1 &= 5-x^2 \\ \Longleftrightarrow 2x^2 - 2x -4 &= 0 \\ \Longleftrightarrow x^2 -x -2 &= 0 \\ \Longleftrightarrow \left(x-\frac{1}{2}\right)^2 -\frac{1}{4} -2 &= 0 \\ \Longleftrightarrow \left(x-\frac{1}{2}\right)^2 &= \frac{9}{4} \\ \Longleftrightarrow x-\frac{1}{2} &= \pm \frac{3}{2} \\ \Longleftrightarrow x &= \frac{1}{2}\pm \frac{3}{2} =2,\, -1. \end{align*}$

$\noindent So the two curves intersect at $x=-1$ and $x=2$. So the area $A$ is$

$\begin{align*}A&= \int _{-1}^{2} \left(5-x^2 - \left(x-1\right)^2 \right) \text{ d}x \\ &= \int _{-1}^{2} \left(5-x^2 - x^2 + 2x -1 \right) \text{ d}x \\ &= \int _{-1}^{2} \left(-2x^2+2x + 4 \right) \text{ d}x \\ &= \left[\frac{-2x^3}{3} +x^2 + 4x \right]_{-1} ^{2} \\ &= \left(\frac{-2(2)^3}{3} +(2)^2 + 4(2) \right) -\left(\frac{-2(-1)^3}{3} +(-1)^2 + 4(-1) \right) \\ &= 9.\end{align*}$

$\noindent Hence the area is 9 square units.$

excellence1 · Feb 7, 2016

Thank you so much. Just noticed my silly/ stupid mistake

Need help with integration (1 Viewer)

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