# Area under the curve - Integrals (Antiderivatives)

I have a question regarding antiderivatives and area under the curve.

I've learned that first, you must to a graph to see if the area is above or below the curve. If it is above the $$x$$-axis the area is "positive" and I must use $$A=\int f(x) dx$$. If it is below the $$x$$-axis the area is "negative" and I must use $$A=-\int f(x) dx$$. In this last one, I've understood that negative outside the integral is because the integration alone will be negative because is under $$x$$-axis, but an area can't be negative so that's why is multiply by that negative. I've also seen this with absolute value $$A=|\int f(x) dx|$$ that I think have the same purpose.

This is an example of an exercise:

Determine the area of the region bounded by the curve of the function $$f(x)=4x^3-16x$$ the $$x$$-axis and the lines $$x=-2$$ y $$x=2$$.

Ok. I'll show you my work.

I first do the graph.

I see that between $$-2$$ and $$0$$ the region bounded is above the $$x$$-axis so is positive, and that between $$0$$ and $$2$$ the region bounded is below the $$x$$-axis so is negative. So I'll call the first one $$A_1$$ and the second $$A_2$$.

$$A_{total}=\int_{-2}^2 (4x^3-16x) dx$$ $$A_{total}=A_1+A_2$$ $$A_{total}=\int_{-2}^0 f(x) dx+(-\int_0^2 f(x) dx)$$ $$A_{total}=\int_{-2}^0 (4x^3-16x) dx+(-\int_0^2 (4x^3-16x) dx)$$ $$A_{total}=[\frac{4x^4}{4}-\frac{16x^2}{2}]|^{0}_{-2} - [\frac{4x^4}{4}-\frac{16x^2}{2}]|^{2}_{0}$$ $$A_{total}=[x^4-8x^2]|^{0}_{-2} - [x^4-8x^2]|^{2}_{0}$$ $$A_{total}=[((0)^4-8(0)^2)-((-2)^4-8(-2)^2)]-[((2)^4-8(2)^2)-((0)^4-8(0)^2)]$$ $$A_{total}=[-(16-32)]-[16-32]$$ $$A_{total}=[-(-16)]-[-16]$$ $$A_{total}=16+16$$ $$A_{total}=32u^2$$

So I got that the total area is 32 square units. But I was wondering why is this different from doing the integration of $$\int_{-2}^2 (4x^3-16x) dx$$. This gives $$0$$.

$$\int_{-2}^2 (4x^3-16x) dx$$ $$=[\frac{4x^4}{4}-\frac{16x^2}{2}]|^{2}_{-2}$$ $$=[x^4-8x^2]|^{2}_{-2}$$ $$=[(2)^4-8(2)^2]-[(-2)^4-8(-2)^2]$$ $$=[16-32]-[16-32]$$ $$=-16-[-16]$$ $$=-16+16$$ $$=0$$

So I'm a bit confused. Which is one correct?

• Note that $|\int f(x)dx|\ne \int |f(x)|dx$ – Henry Lee May 12 at 22:30

Ok, so here's the cause of your confusion: the meaning of the word "area" depends on the context of the problem. Indeed your calculation

$$\int_{-2}^{2} (4x^3-16x)\,dx = 0$$

is correct. (A shortcut is to notice that you are integrating an odd function over a domain which is symmetric about the origin). However, in this particular question, the "area between the curves and the x-axis" really means to compute

$$\int_{-2}^{2} |4x^3-16x| \,dx$$

So yes, in the geometric sense, area generally must be positive. But when evaluating definite integrals, we sometimes think of the area above the x-axis as being "positive area" and the area below the x-axis as "negative area."

$$\int_{-2}^{2} |4x^3-16x|\,dx = 2 \cdot \int_{-2}^{0} (4x^3-16x)\,dx = 2 \big[x^4 - 8x^2\big] \big|_{-2}^{0} = 32.$$

• (+1) A very good answer for a new contributor! Welcome to MSE! – clathratus May 13 at 3:47

In such cases , you must create the partial intervals and integrate over them.

The integral without considering whether the function is negative or positive counts the areas above the $$x$$-axis positive and the areas below the $$x$$-axis negative.

Hence, the solution $$0$$ is not correct (anyway area $$0$$ cannot be the result here).

You need the roots of the function to find the necessary intervals , exactly what you did in the first (correct) approach.

In your first approach, you are taking the absolute value of the "signed areas" (so the areas will be positive whether above or below the $$x$$-axis). Hence, $$A_1 = \vert 16\vert = 16$$ and $$A_2 = \vert -16\vert = 16$$, so $$A_1+A_2 = 32$$.

When taking the definite integral, you are taking the signs into account, so $$A_1 = 16$$ while $$A_2 = -16$$, which means $$\int_{-2}^2 f(x)dx = 0$$. (The areas above and below the $$x$$-axis are equal and cancel out.)

The first approach is correct because the question is asking for the area bounded by $$f(x)$$ and the $$x$$-axis (in which case, a non-positive area makes no sense), not for the definite integral of $$f(x)$$ between $$-2$$ and $$2$$.