# compute area under two curve

suppose we are give task to calculate area of figure,which is bounded by two curve $y=[x]$ and $y=(2-x^2)$, here $[x]$ denotes modulus,not ceiling or rounding of x.

i use wolframalpha to figure out what kind of location,intersection points has this two figure,here is link of this http://www.wolframalpha.com/input/?i=abs%28x%29%3D2-x%5E2 i see that points of intersection are $-1$ and $1$,also i know that area under two curve $y=f_1(x)$ and $y=f_2(x)$ and intersection points are $x_1$ and $x_2$ is $$\int_{x_1}^{x_2}(f_2(x)-f_1(x))dx$$ but i am confused if i should take $y=[x]$ directly or consider two interval $[-1..0]$ and $[0...1]$ and use $-x$ and $x$ for each interval? please give me some hint

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Do it separately, it is safer. Anyway, you only need to deal with the first quadrant part. By symmetry for the full area you double the result.

Because the curve $y=2-x^2$ is above the curve $y=x$ in the region of interest, the first quadrant part has area $\int_0^1((2-x^2)-x)\,dx$.

I would somewhat prefer to draw the vertical line from the point of intersection $(1,1)$ to the $x$-axis. Then the desired area is $\int_0^1 (2-x^2)\,dx -\int_0^1 x\,dx$. It feels more concrete: area under $y=2-x^2$, above $x$-axis, from $0$ to $1$, minus the area of the "hole."

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so calculate area under 0..1 and the multiply by two yes? – dato datuashvili Jun 18 '12 at 6:55
Yes.${}{}{}{}{}{}$ – André Nicolas Jun 18 '12 at 6:56
i have calculated area in first quadrant,get $2/3$,so then i have multiplyed it by 2,so get $4/3$ – dato datuashvili Jun 18 '12 at 6:58
thanks very much for detailed answer – dato datuashvili Jun 18 '12 at 7:01
Really? I get a different number. – André Nicolas Jun 18 '12 at 7:01

the absolute value of x is an even function (Y1=[x]) and (Y2=2-x^2) is also a even function therefor y2-y1 is an even function by consequence.then the integral of y2-y1 between -1 and 1 is equal to the integral of Y2-Y1 enter 0 and 1 multiplied by two. P.S: In the interval [0 1] [x]=x. area of figure = 2* 3.5/3=7/3.

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thanks you for help.+1 – dato datuashvili Jun 18 '12 at 15:33