넓이(여러 가지 함수)

If the curves $y=f(x)=x$ and $y=g(x)=3x-x^2$ intersect at $(x,y)$, then

$$\begin{aligned} 3x-x^2&=y=x\\ x^2-2x&=0\\ x(x-2)&=0\\ x=0 \quad &\text{or} \quad x=2 \end{aligned}$$

Furthermore, $g(x)-f(x) = 2x-x^2 = x(2-x)$ is positive for all $0\le x\le 2$. That is, the curve $y=3x-x^2$ lies above the line $y=x$ for all $0\le x\le 2$.

We therefore evaluate the integral:

$$\begin{aligned} \int_0^2 \big[ (3x-x^2) - x \big] \,\mathrm{d}x = \int_0^2 [2x-x^2]\,\mathrm{d}x = \bigg[x^2 - \frac{x^3}{3}\bigg]^{2}_{0} = \bigg[ 4-\frac{8}{3} \bigg] -0 = \frac{4}{3} \end{aligned}$$

For our computation, we will need an antiderivative of $x^2\sqrt{x^3+1}$, which can be found using the substitution $u=x^3+1$, $\mathrm{d}u = 3x^2\,\mathrm{d}x$:

$$\begin{aligned} \int x^2\sqrt{x^3+1} \, \mathrm{d}x = \int \sqrt u \cdot \frac13\,\mathrm{d}u = \frac13\int u^{1/2}\,\mathrm{d}u = \frac13\cdot \frac{u^{3/2}}{3/2}+C = \frac29(x^3+1)^{3/2} + C. \end{aligned}$$

The two functions $f(x)$ and $g(x)$ are clearly equal at $x=0$. If $x\ne0$, then the functions are equal when

$$\begin{aligned} 3x^2 &= x^2\sqrt{x^3+1} \\ 3 &= \sqrt{x^3+1} \\ 9 &= x^3+1 \\ 8 &= x^3 \\ 2 &= x. \end{aligned}$$

The function $g(x)=3x^2$ is the larger of the two on the interval $[0,2]$, as can be seen by plugging in $x=1$, say, or by observing that when $x$ is very small $f(x)=x^2\sqrt{x^3+1}\approx x^2$ and $g(x)=3x^2$.

The area in question is therefore:

$$\begin{aligned} \int_0^2 \big( 3x^2 - x^2\sqrt{x^3+1} \big) \, \mathrm{d}x &= \bigg( {x^3} - \frac29(x^3+1)^{3/2} \bigg) \bigg|_0^2 \\ &= \bigg( 2^3 - \frac2 9(2^3+1)^{3/2} \bigg) - \bigg( 0^3 - \frac29(0^3+1)^{3/2} \bigg) \\ &= \bigg( 8 - 6 \bigg) - \bigg( 0 - \frac 2 9 \bigg) =\frac{20}9. \end{aligned}$$

Let's begin by sketching our region. Note that $y=\sqrt{1-x^2}$ and $y=\sqrt{9-x^2}$ are the top halves of circles centred at the origin with radii 1 and 3, respectively.

Our region is the difference of two quarter-circles, so we find its area using geometry: $$\text{Area}=\frac{1}{4}\left(\pi\cdot 3^2\right)-\frac{1}{4}\left(\pi\cdot 1^2\right)=2\pi$$