Find the area enclosed by the curves: $$ y=x^2;\quad y^2=2x-x^2$$
I know how to set-up the problem. I am having difficulty figuring out the integral however. I first found the points of intersection by setting the two equations equal to one another. I got the lower limit to be $0$ and the upper limit to be $1$. Since the second equation is higher than the first, I set up the integral so that the first is being subtracted from the second. Hence, the integral looked like $$\int_{0}^{1} \left(\sqrt{2x-x^2}-x^2\right)\,dx $$ I can't figure out how to integrate that so I can not go further with the problem.
Best Answer
You're set up is spot on. Indeed, we're finding the area bounded below the curve $y = \sqrt{2x - x^2}\,$ and above the curve $y = x^2,\;$ between $x = 0$ and $x = 1$:
So indeed, we need to integrate:
$$\begin{align} I & = \int_0^1 \left(\sqrt{2x - x^2} - x^2\right)\,dx \\ \\ & = \int_0^1 \sqrt{1 - 1 + 2x - x^2} \,dx - \int_0^1 x^2\,dx\\ \\ & = \int_0^1 \sqrt{1 - (x - 1)^2} \,dx - \int_0^1 x^2\,dx \end{align}$$
Now, for the first integral, we use the trigonometric substitution $(x - 1) = \sin\theta$, so that $dx = \cos\theta\,d\theta$.
Finding the new bounds of integration for the first integral (so we can save ourselves the task of "back substitution" at the very end):
At $x = 0, \sin\theta = -1 \implies \theta = -\pi/2.\;$ At $x = 1, \sin\theta = 0 \implies \theta = 0$.
This gives you, after substituting for the first integral:
$$I = \int_{-\pi/2}^{0} \sqrt{1 - \sin^2 \theta}\cos\theta\,d\theta - \int_0^1 x^2\,dx $$
We will use the identities $$\begin{align}\;1 - \sin^2\theta & = \cos^2 \theta\tag{1} \\ \cos^2 \theta & = \dfrac {1 + \cos (2\theta)}{2}\tag{2}\end{align}$$
$$I =\int_{-\pi/2}^0 \left(\sqrt{\cos^2\theta}\right)\cos\theta\,d\theta = \int_{-\pi/2}^0 \cos^2\theta\,d\theta - \int_0^1 x^2\,dx \tag{1}$$
$$I = \dfrac 12\int_{-\pi/2}^0 \left(1 + \cos(2\theta)\right) \,d\theta - \int_0^1 x^2 \,dx\tag{2}$$
Integration should now be relatively straightforward:
$$I = \left[\dfrac \theta2 + \dfrac 14\sin(2\theta)\right]\Big|_{-\pi/2}^0 \;- \;\dfrac{x^3}3\Big|_0^1\quad = \quad \frac\pi4-\dfrac 13$$