# Finding the area under the curve from a graph

What is the area of the shared region in the figure above that is bounded by the $$x$$-axis and the curve with the equation $$y=x\sqrt{1-x^2}$$?

This is the problem I was given. I assumed the answer was $$0$$ - the positive and negative areas should cancel each other out. That answer was incorrect. I then thought the answer might be $$\frac13$$ and they are only asking for the area above the $$x$$-axis. That was also incorrect. The answer that was given as correct was: $$\frac23$$. I assume that is because it is the area of the top - which is $$\frac13$$ and the area of the bottom also $$\frac13$$ - which makes $$\frac23$$. But, why is the answer not zero? Doesn't integration count area under the $$x$$-axis as negative? Does the wording of the question say otherwise?

• Why do you think an area can be negative? That's like saying that the distance between the points $4$ and $3$ is $-1$ beacuse "you go back". The area of a curve is the integral of the absolute value of the function. – Alfredo Aug 1 '19 at 20:42

In your question, it asks for the area of the shaded region, and area is always positive.

For integration, it is taught as the area between the curve and the $$x$$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $$x$$-axis, and offset by those below $$x$$-axis).

Note that Area is absolute, but Net Area is relative

The integral $$\int_{-1}^0y~dx$$ is negative, but the area is always non-negative. So the area of the region is $$\int_0^1y~dx-\int_{-1}^0y~dx$$.

The function is odd, the area you look for is $$A=2\int_0^1x\sqrt{1-x^2}dx$$

put $$y=x^2$$.

then

$$A=\int_0^1\sqrt{1-y}dy$$ $$=\int_0^1(1-y)^{\frac 12}dy$$

$$=\Bigl[ \frac 23(1-y)^{\frac 32}\Bigr]_1^0$$ $$=\frac 23.$$

Let $$x=\sin t$$.

Thus, we need to get $$2\int\limits_0^1x\sqrt{1-x^2}dx=2\int\limits_0^{\frac{\pi}{2}}\sin{t}\cos^2tdt=\int\limits_0^{\frac{\pi}{2}}\sin2t\cos{t}=$$ $$=\frac{1}{2}\int\limits_0^{\frac{\pi}{2}}(\sin3t+\sin{t})dt=-\frac{1}{6}\cos3t-\frac{1}{2}\cos{t}\big{|}_0^{\frac{\pi}{2}}=\frac{2}{3}.$$