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The exercise is: calculate the area of the plane region limited by the curves $$y= x^2$$ $$y=x^2/2$$ $$y=x$$

I know that I need to use integrals and I know how to apply them, my only difficulty in this problem is that I'm having trouble by finding what is the figure on the plane...

Shailesh
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Granger Obliviate
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1 Answers1

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To help you discover the points that may interest you, try setting these equations equal to each other and see what results.

$$x^2=x^2/2\Rightarrow x=0$$ $$x^2=x\Rightarrow x=0\text{ or }x=1$$ $$x^2/2=x\Rightarrow x=0\text{ or }x=2$$

So our points of interest are clustered in $[0,2]$. You will then want to accurately draw the three functions on that domain, and it will become evident what the area is. Here's a generated picture so that you may check the accuracy of your drawing.

Functions on [0,2]

For reference, the blue curve is $x^2$, the green curve is $x$, and the orange curve is $x^2/2$.

There is an area bounded solely by $x$ and $x^2$, however we desire the area directly right of that, bounded above by $x^2$ and then $x$ and below by $x^2/2$. This area can be calculated by

$$A=\int_0^1(x^2-\frac{x^2}{2})\,dx+\int_1^2(x-\frac{x^2}{2})\, dx=\frac{1}{2}.$$

superckl
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