넓이(여러 가지 함수)

We need to find where the curves intersect.

$$\begin{aligned} \frac{x^2}{4}=y^2&=6-\dfrac{5x}{4}\\ \frac{1}{4}x^2+\frac{5}{4}x-6&=0\\ x^2+5x-24&=0\\ (x+8)(x-3)&=0\\ x=-8,\quad x&=3 \end{aligned}$$

The curves intersect at $(-8,4)$ and $(3,-\frac{3}{2})$. Using horizontal strips:

we have

$$\begin{aligned} \text{Area} = \int_{-3/2}^{4}\Big[\frac{4}{5}(6-y^2)+2y\Big]\ \mathrm{d}y \end{aligned}$$

If the curves intersect at $(x,y)$, then

$$\begin{aligned} \left(x^2\right)^2&=\left(4a\right)^2y^2 = (4a)^24ax\\ x^4&=(4a)^3 x\\ x^4&-(4a)^3x=0\\ x(&x^3-(4a)^3)=0\\ x&= 0 \quad\text{or}\quad x^3=(4a)^3 \end{aligned}$$

The curves intersect at $(0,0)$ and $(4a,4a)$. (It is also possible to find these points by inspection.) Using vertical strips:

We want the $y$-values of the functions. We write the top function as $y =\sqrt{4ax}$ (we care about the positive square root, not the negative one) and we write the bottom function as $y=\frac{x^2}{4a}$. Then we have

$$\begin{aligned} \text{Area} = \int_0^{4a}\left[\sqrt{4ax}-\frac{x^2}{4a}\right]\ \mathrm{d}x \end{aligned}$$

The curves intersect when $x=4y^2$ and $0=4y^2+12y+5 =(2y+5)(2y+1)$. So, the curves intersect at $(1,-\tfrac{1}{2})$ and $(25,-\frac{5}{2})$. Using vertical strips:

we have

$$\begin{aligned} \text{Area} = \int_1^{25}\left[-\frac{1}{12}(x+5)+\frac{1}{2}\sqrt{x}\right]\ \mathrm{d}x \end{aligned}$$