넓이(여러 가지 함수)

We need to figure out which curve is on top, when. To do this, set $h(x) = 3x - x\sqrt{25-x^2}$. If $h(x) > 0$, then $y=3x$ is the top curve; if $h(x)<0$, then $y=x\sqrt{25-x^2}$ is the top curve.

$$\begin{aligned} h(x) &= 3x - x\sqrt{25-x^2} = x\left[3-\sqrt{25-x^2}\right] \\ \text{We only care about values of $x$ in $[0,4]$, so $x$ is nonnegative. Then $h(x)$ is positive when:} \\ 3&> \sqrt{25-x^2}\\ 9&> 25-x^2\\ x^2 & > 16\\ x&> 4 \end{aligned}$$

That is, $h(x)$ is never positive over the interval $[0,4]$. So, $y = x \sqrt{25-x^2}$ lies above $y=3x$ for all $0\le x\le 4$.

The area we need to calculate is therefore:

$$\begin{aligned} A &= \int_0^4 \left[x \sqrt{25-x^2} - 3x\right]\,\mathrm{d}x \\ &= \int_0^4 x \sqrt{25-x^2}\,\mathrm{d}x - \int_0^4 3x\,\mathrm{d}x \\&= A_1 - A_2. \end{aligned}$$

To evaluate $A_1$, we use the substitution $u(x) = 25-x^2$, for which $\mathrm{d}u = u'(x)\,\mathrm{d}x= -2x\,\mathrm{d}x$; and $u(4)=25-4^2=9$ when $x=4$, while $u(0)=25-0^2=25$ when $x=0$. Therefore

$$\begin{aligned} A_1 &= \int_{x=0}^{x=4} x \sqrt{25-x^2}\,\mathrm{d}x = -\frac{1}{2} \int_{u=25}^{u=9} \sqrt{u}\,\mathrm{d}u = \left[-\frac{1}{3} u^{3/2} \right]_{25}^{9} = \frac{125 - 27}{3} = \frac{98}{3} \end{aligned}$$

For $A_2$ we use the antiderivative directly:

$$A_2 = \int_0^4 3x\,\mathrm{d}x =\left[ \frac{3x^2}{2} \right]_0^4 = 24$$

Therefore the total area is:

$$\begin{aligned} A = \frac{98}{3} - 24 = \frac{26}{3} \end{aligned}$$