# Area under the curve $f(x)=x^3$

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.

• what are you doing? Oct 19 '19 at 15:55
• And why do you think you are doing something wrong?
– smcc
Oct 19 '19 at 15:56
• I thought that both values should match Oct 19 '19 at 15:59
• 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. Oct 19 '19 at 15:59
• @Bernard thanks . Oct 19 '19 at 16:05

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 $$\int_{-1}^1x^3dx=\int_{0}^1x^3dx+\int_{-1}^0x^3dx=\frac14-\frac14=0.$$

• 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? Oct 19 '19 at 16:01
• I think that you should ask your professor Oct 19 '19 at 16:11

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.

• $$\int f(x)\mathop{dx}$$ 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 $$> 0.$$