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Find the area of ​​the figure that is bound by the lines:

$y=|x^2-1|$ and $y=3|x|-3$

I tried: $$A=\int_{-1}^{1}(|x^2-1|-3|x|+3)dx$$

Is it correct? Please, help for the right solution.

I dont know, how can I evaluate $\int|x|dx.$

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    $\begingroup$ Naturally, split the integral into $\int_{-1}^0$ and $\int_0^1$ and integrate, knowing what $||$ is on either side. Alternately, you may note that the function is even, so $\int_{-1}^1 (|x^2 -1| - 3|x| + 3) dx = 2 \int_0^1 (|x^2 - 1| - 3|x| + 3) dx$, and now can you remove the modulus $\endgroup$ Commented Jul 19, 2018 at 3:58
  • $\begingroup$ Student, there are more regions bounded by the two curves and to evaluate $\int_{-1}^{1} |x|$ split it in two integrals i.e. $\int_{-1}^{1} |x|=\int_{-1}^{0} -x+\int_{0}^{1} x$ $\endgroup$
    – MathBS
    Commented Jul 19, 2018 at 14:52

3 Answers 3

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You should draw the graphs of the two given curves carefully before solving this type of problems, and to do that you have two find the points at which they intersect each other. Let's do that
First write the equation of two given curves explicitly-
$ y=|x^2-1|= \begin{cases} x^2-1, & \text{if $x\in(-\infty,-1]\cup[1,\infty)$}\\ 1-x^2, & \text{if $x\in(-1,1)$} \end{cases} $
$ y=3|x|-3= \begin{cases} 3x-3, & \text{if $x\ge0$}\\ -3x-3, & \text{if $x<0$} \end{cases} $
Now, find the points at which the cut each other.
Case 1: $x\ge1$
The equation of two curves become
$y=x^2-1$ and $y=3x-3$
A simple calculation shows that these two curves intersects at the points $(1,0)$ and $(2,3)$ in $[1,\infty)$.
Case 2: $x\in[0,1)$
The equation of two curves become
$y=1-x^2$ and $y=3x-3$
These two curves don't intersect each other in $[0,1)$
Case 3: $x\in(-1,0)$
The equation of two curves become
$y=1-x^2$ and $y=-3x-3$
These two curves don't intersect each other in $(-1,0)$
Case 4: $x\le-1$
The equation of two curves become
$y=x^2-1$ and $y=-3x-3$
A simple calculation shows that these two curves intersects at the points $(-1,0)$ and $(-2,3)$ in $(-\infty,-1]$.
So, combining all the cases we get, two curves intersect each other at the points $A(-2,3)$, $B(-1,0)$, $C(1,0)$ and $D(2,3)$.
Note one thing the curves $y=|x^2-1|$, $y=3|x|-3$ intersect $Y$-axis at $E(0,1)$, $F(0,-3)$ respectively.
enter image description here
So, the area of the region bounded by these two curves on the plane=
Area of the region $(ABA+BCEB+\Delta BFC+CDC)=$
$ =\int_{-2}^{-1} \lbrace(-3x-3)-(x^2-1)\rbrace dx\quad+\int_{-1}^{1} (1-x^2) dx\quad+\frac{1}{2} \times 2 \times 3\quad+\int_{1}^{2} \lbrace(3x-3)-(x^2-1)\rbrace dx $
$ =2\int_{1}^{2} \lbrace(3x-3)-(x^2-1)\rbrace dx\quad+2\int_{0}^{1} (1-x^2) dx\quad+3 $
$ =\frac{1}{3}+\frac{4}{3}+3=\frac{14}{3} $
So, the area of the region bounded by the curves is $\frac{14}{3}$ units.

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  • $\begingroup$ Thank you very much, is csch2's answers wrong?? $\endgroup$ Commented Jul 19, 2018 at 14:58
  • $\begingroup$ What is $...dx×2x...$ ?? $\endgroup$ Commented Jul 19, 2018 at 15:14
  • $\begingroup$ Sorry, it's my typing mistake, it is product sign, look I'm calculating the area of $\Delta BFC$ using the general formula for finding the area of triangle 1/2 x base x height $\endgroup$
    – MathBS
    Commented Jul 19, 2018 at 18:02
  • $\begingroup$ csch2, and you both are neglecting the area in $(-2,-1)$ and $(1,2)$. Is my solution okay for you? I have tried to represent the solution as elaborate and simple as possible. $\endgroup$
    – MathBS
    Commented Jul 19, 2018 at 18:04
  • $\begingroup$ I have edited the ambiguity of the product sign $\endgroup$
    – MathBS
    Commented Jul 19, 2018 at 18:14
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Note that the area bounded by the equations is between $x=-1$ and $x=1$ and that $\left|x^2-1\right|=1-x^2$ on the interval $(-1,1)$. Also note that $3\left|x\right|-3=-3x-3$ for $x<0$ and $3x-3$ for $x\geq0$. Then your integrals become:

$$\int_{-1}^0\left(1-x^2\right)-\left(-3x-3\right)\text dx+\int_0^1\left(1-x^2\right)-\left(3x-3\right)\text dx$$ $$=\int_{-1}^0-x^2+3x+4\text dx+\int_0^1-x^2-3x+4\text dx$$ which are simple to evaluate. Alternatively, as астон вілла олоф мэллбэрг mentioned, you could note that the area has even symmetry and simply double one of the integrals, which will give you the same area.

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  • $\begingroup$ csch2, you are missing the region in between $(-2,-1)$ and $(1,2)$ $\endgroup$
    – MathBS
    Commented Jul 19, 2018 at 14:56
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HINT.- By clear symmetrie you have the required area is equal to twice $$\int_0^1(x^2-3x+2)dx+\int_1^2(3x-x^2-2)dx$$

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  • $\begingroup$ There are $4$ area, isn't it? $\endgroup$ Commented Jul 19, 2018 at 15:39
  • $\begingroup$ Yes it is. For this the two integrals taken twice. $\endgroup$
    – Piquito
    Commented Jul 19, 2018 at 15:52
  • $\begingroup$ Please, can you edit more details.. thank you.??(+1) $\endgroup$ Commented Jul 19, 2018 at 15:55
  • $\begingroup$ The two integrals are taken for $x\gt 0$ which is enough by symmetrie. Look at the Biswarup Saha's answer for understand in his graph the change of sign in the second integral. $\endgroup$
    – Piquito
    Commented Jul 19, 2018 at 16:18
  • $\begingroup$ Look just at the graph and nothing else. $\endgroup$
    – Piquito
    Commented Jul 19, 2018 at 16:41

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