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If $f(x)=\min\left\{{\sqrt{1-x^2 /4},e^{-x}}\right\}$ where x belongs to [-2,2]. We have to find the area between the curve and $x$-axis .

I found the area between $(-2,0)$ to $(0,0)$.

But in the first quadrant I am unable to find their intersection point.

user135482
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  • How to solve this without using calculator – user135482 Apr 24 '17 at 14:03
  • There is something missing. What is $(-2,)$? And what do you mean by this anyway? Your function seems to be defined on $\mathbb{R}$ so you can't find the area between two points if that is what you mean by $(a,b)$. Furthermore what do you mean by intersection point? – Jakob Elias Apr 24 '17 at 14:09
  • Consider asking your questions more precisely in the future. – Jakob Elias Apr 24 '17 at 14:10
  • @JakobElias sorry , I have edited it – user135482 Apr 24 '17 at 14:11
  • I suspect that the OP simply meant "between x= -2 and x= 0", the points on x-axis (-2, 0) and (0, 0). It is easy to see that x= 0, y= 1 is one intersection point. There is no intersection point for x< 0 so the integral would just be from x= -2 to x= 0. – user247327 Apr 24 '17 at 14:15

1 Answers1

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Using

$$ \min[f(x),g(x)] = \frac{f(x)+g(x)}{2} - \frac{|f(x)+g(x)|}{2} ~, $$ your function becomes $$ f(x)=\min\left\{{\sqrt{1-x^2 /4},e^{-x}}\right\} = \frac{\sqrt{1-x^2 /4}+e^{-x}}{2} - \frac{|\sqrt{1-x^2 /4}+e^{-x}|}{2} ~. $$

The area between your function and the x-axis on the interval $[-2,2]$ is defined by the definite integral between $x=-2$ and $x=2$. Therefore evaluating the integral

$$ \text{Area} = \int_{-2}^{2}f(x)dx ~, $$ will give you the area you are looking for.

tpvasconcelos
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