정적분의 뜻과 성질

(a) Since $f(x)$ is positive on $[-1,1]$, the area bounded by the function and $x$-axis on the given interval is $A_1 = \int_{-1}^{1} (3-x^2) \, dx$. A sketch of the function and region is shown below. (b) Likewise, because $g(x)$ is positive on $[-1,1]$, the area bounded by the function and $x$-axis on the given interval is $A_2 = \int_{-1}^{1} 2x^2 \, dx$. A sketch of the function and region is shown below. (c) From the above two figures, we see that the exact area between the two curves is the area under $f$ ($A_1$) minus the area under $g$ ($A_2$). That is, $$A_1 - A_2 = \int_{-1}^{1} (3-x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx\,.$$ (d) By the sum rule for definite integrals over the same interval, we know that $$\int_{-1}^{1} (3-x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx = \int_{-1}^{1} [ (3-x^2) - 2x^2 ] \, dx$$ and thus the latter single integral captures the exact area between the two curves on $[-1,1]$. (e) In general, if $p(x) \ge q(x)$ on $[a,b]$, the exact area, $A$, between $p$ and $q$ on the interval will be given by $$A = \int_a^b [p(x) - q(x)] \, dx$$ because we can find the area as the difference of the area under $p$ ($\int_a^b p(x) \, dx$) and the area under $q$ ($\int_a^b q(x) \, dx$), and then subtract the two definite integrals as shown in (d).