넓이(여러 가지 함수)

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:

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