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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 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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up vote 2 down vote accepted

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

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