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Since $f(x)=x^3$ is an odd function so integrating it from $-1$ to $1$ will be zero by the property of definite integral. If we integrate this function from $0$ to $1$ we get $1/4$ and from $-1$ to $0$ we get $-1/4$. Area can’t be negative so taking absolute value of both parts we get $1/2$. What am I doing wrong? Please explain it.

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    $\begingroup$ what are you doing? $\endgroup$
    – Azlif
    Oct 19 '19 at 15:55
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    $\begingroup$ And why do you think you are doing something wrong? $\endgroup$
    – smcc
    Oct 19 '19 at 15:56
  • $\begingroup$ I thought that both values should match $\endgroup$
    – Nancy
    Oct 19 '19 at 15:59
  • $\begingroup$ The integral yields an algebraic area, i.e. an area with a sign. If you want the areain the geometric sense, you're doing exactly what has to be done. $\endgroup$
    – Bernard
    Oct 19 '19 at 15:59
  • $\begingroup$ @Bernard thanks . $\endgroup$
    – Nancy
    Oct 19 '19 at 16:05
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As you know the definite integral doesn't find the area, but the signed area, in the sense that regions below the $x$-axis count negatively towards the value of the integral. The area between the graph of $y=x^3$ and the $x$-axis between $x=-1$ and $x=1$ is given by $$\left|\int_0^1x^3dx\right|+\left|\int_{-1}^0x^3dx\right|=\frac14+\frac14=\frac12,$$ but the total signed area is given by $$\int_{-1}^1x^3dx=\int_{0}^1x^3dx+\int_{-1}^0x^3dx=\frac14-\frac14=0.$$

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  • $\begingroup$ So unless it is mentioned that we have to find signed area , it means we just need to find area bounded by the curve by taking absolute value? $\endgroup$
    – Nancy
    Oct 19 '19 at 16:01
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    $\begingroup$ I think that you should ask your professor $\endgroup$ Oct 19 '19 at 16:11
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If I don't misunderstand your words, I believe that you misunderstand the meaning of integrating a function, which leads to the signed area under a function, instead of the area under a function. More specifically speaking, signed area means that the area below the x-axis is negative and the area above the x-axis is positive.

Hope this can help you.

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  • $\int f(x)\mathop{dx}$ gives the nett signed area between the curve $y=f(x)$ and the $x$-axis.

    (Signed areas below the $x$-axis are negative.)

    An odd function $f$ is symmetric with respect to the origin, so its signed area in the interval $[-a,a]$ is $\displaystyle\int_{-a}^{a} f(x)\mathop{dx}$ $$=\,-\left(\int_{0}^{a}f(x)\mathop{dx}\right)+\int_{0}^{a} f(x)\mathop{dx}\,=\,0.$$

  • $\int \lvert f(x)\rvert \mathop{dx}$ gives the nett (unsigned) area between the curve $y=f(x)$ and the $x$-axis.

    An odd function $f$ is congruent left and right of the $y$-axis, so its area in the interval $[-a,a]$ is $\displaystyle\int_{-a}^{a} \lvert f(x)\rvert\mathop{dx}$ $$=\,2\int_{0}^{a} \lvert f(x)\rvert\mathop{dx},$$ which is generally $> 0.$

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