[월:] 2025년 04월

곡선 \(y=x^3\)과 \(y=2x-x^2\)으로 둘러싸인 도형의 넓이를 구하라.

풀이

교점의 \(x\) 좌표를 구한다.

\(x^3=2x-x^2\)

\(x^3+x^2-2x=0\)

\(x(x^2+x-2)=0\)

\(x(x+2)(x-1)=0\)

\(x=-2,0,1\)

적분구간 \(-2\leq x\leq0\)에서 \(x^3\geq 2x-x^2\)이고, 적분구간 \(0\leq x\leq1\)에서 \(x^3\leq 2x-x^2\)이다.

도형의 넓이는

\(\displaystyle\int_{-2}^0(x^3-(2x-x^2))dx+\int_{0}^1(2x-x^2-x^3)dx\)

\(\displaystyle=\int_{-2}^0(x^3+x^2-2x)dx+\int_{0}^1(-x^3-x^2+2x)dx\)

\(\displaystyle=\left[\frac{x^4}{4}+\frac{x^3}{3}-x^2\right]_{-2}^0+\left[-\frac{x^4}{4}-\frac{x^3}{3}+x^2\right]_{0}^1\)

\(\displaystyle=0-\left(\frac{(-2)^4}{4}+\frac{(-2)^3}{3}-(-2)^2\right)+\left(-\frac{1^4}{4}-\frac{1^3}{3}+1^2\right)-0\)

\(\displaystyle=-\left(4-\frac{8}{3}-4\right)+\left(-\frac{1}{4}-\frac{1}{3}+1\right)\)

\(\displaystyle=\frac{8}{3}+\frac{12-3-4}{12}=\frac{32+5}{12}=\frac{37}{12}\)

이다.

두 포물선 \(y=x^2+3x\)와 \(y=-2x^2+6\)으로 둘러싸인 도형의 넓이를 구하라.

풀이

교점의 \(x\) 좌표를 구한다.

\(x^2+3x=-2x^2+6\)

\(3x^2+3x-6=0\)

\(x^2+x-2=0\)

\((x+2)(x-1)=0\)

\(x=-2,1\)

따라서 적분구간은 \(-2\leq x\leq1\)이 되고, 이 구간에서 \(-2x^2+6\geq x^2+3x\)이다.

도형의 넓이는

\(\displaystyle\int_{-2}^1(-2x^2+6-x^2-3x)dx=\int_{-2}^1(-3x^2-3x+6)dx\)

\(\displaystyle=\left[-x^3-\frac{3x^2}{2}+6x\right]_{-2}^1\)

\(\displaystyle=-1-\frac{3}{2}+6-\left(-(-2)^3-\frac{3(-2)^2}{2}+6(-2)\right)\)

\(\displaystyle=\frac{7}{2}-(8-6-12)=\frac{7}{2}+10=\frac{27}{2}\)

이다.

\(\displaystyle f(x)=\left\{\begin{array}{cl}x,&-2\leq x<1\\x^2,&1\leq x<2\\(x-4)^2,&2\leq x\leq3\end{array}\right.\)일 때 \(\displaystyle\int_{-2}^3 f(x)dx\)를 구하라.

풀이

\(\displaystyle\int_{-2}^3 f(x)dx=\int_{-2}^1 xdx+\int_1^2 x^2dx+\int_2^3 (x-4)^2dx\)

\(\displaystyle=\left[\frac{x^2}{2}\right]_{-2}^1+\left[\frac{x^3}{3}\right]_1^2+\left[\frac{(x-4)^3}{3}\right]_2^3\)

\(\displaystyle=\frac{1}{2}-2+\frac{8}{3}-\frac{1}{3}-\frac{1}{3}+\frac{8}{3}=-\frac{3}{2}+\frac{14}{3}\)

\(\displaystyle=\frac{28-9}{6}=\frac{19}{6}\)

다음 적분을 구하라.

$$\int_2^3\frac{t^3}{t+1}dt+\int_2^3\frac{1}{u+1}du$$

풀이

\(\displaystyle\int_2^3\frac{t^3}{t+1}dt+\int_2^3\frac{1}{u+1}du\)

\(\displaystyle=\int_2^3\frac{(t+1)(t^2-t+1)-1}{t+1}dt+\int_2^3\frac{1}{u+1}du\)

\(\displaystyle=\int_2^3\frac{(t+1)(t^2-t+1)}{t+1}dt-\int_2^3\frac{1}{t+1}dt+\int_2^3\frac{1}{u+1}du\)

\(\displaystyle=\int_2^3(t^2-t+1)dt=\left[\frac{t^3}{3}-\frac{t^2}{2}+t\right]_2^3\)

\(\displaystyle=\frac{3^3}{3}-\frac{3^2}{2}+3-\left(\frac{2^3}{3}-\frac{2^2}{2}+2\right)\)

\(\displaystyle=9-\frac{9}{2}+3-\left(\frac{8}{3}-2+2\right)\)

\(\displaystyle=\frac{15}{2}-\frac{8}{3}=\frac{45-16}{6}=\frac{29}{6}\)

다음 적분을 구하라.

$$\int_{-1}^2\vert x-1\vert dx$$

풀이

\(\displaystyle\int_{-1}^2\vert x-1\vert dx=\int_1^2(x-1)dx+\int_{-1}^1(-x+1)dx\)

\(\displaystyle=\left[\frac{1}{2}x^2-x\right]_1^2+\left[-\frac{1}{2}x^2+x\right]_{-1}^1\)

\(\displaystyle=\frac{2^2}{2}-2-\left(\frac{1^2}{2}-1\right)-\frac{1^2}{2}+1-\left(-\frac{(-1)^2}{2}-1\right)\)

\(\displaystyle=0-\left(-\frac{1}{2}\right)+\frac{1}{2}-\left(-\frac{3}{2}\right)=\frac{5}{2}\)

다음 적분을 구하라.

$$\int_1^3\frac{u^4-8}{u^2}du$$

풀이

\(\displaystyle\int_1^3\frac{u^4-8}{u^2}du=\int_1^3\left(u^2-8u^{-2}\right)du\)

\(\displaystyle=\left[\frac{1}{3}u^3-8(-1)u^{-1}\right]_1^3=\left[\frac{u^3}{3}+\frac{8}{u}\right]_1^3\)

\(\displaystyle=\frac{3^3}{3}+\frac{8}{3}-\left(\frac{1^3}{3}+\frac{8}{1}\right)=\frac{35}{3}-\frac{25}{3}=\frac{10}{3}\)

다음 적분을 구하라.

$$\int_0^8\sqrt[3]{x}dx$$

풀이

\(\displaystyle\int_0^8\sqrt[3]{x}dx=\int_0^8x^{\frac{1}{3}}dx=\left[\frac{3}{4}x^{\frac{4}{3}}\right]_0^8=\left[\frac{3}{4}x\sqrt[3]{x}\right]_0^8\)

\(\displaystyle=\frac{3}{4}(8)\sqrt[3]{8}-\frac{3}{4}(0)\sqrt[3]{0}=6\sqrt[3]{2^3}-0=6(2)=12\)

다음 적분을 구하라.

$$\int_1^2\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx$$

풀이

\(\displaystyle\int_1^2\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx=\int_1^2\left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)dx\)

\(\displaystyle=\left[\frac{2}{3}x^{\frac{3}{2}}-2x^{\frac{1}{2}}\right]_1^2=\left[\frac{2}{3}\sqrt{x^3}-2\sqrt{x}\right]_1^2\)

\(\displaystyle=\left[\frac{2}{3}x\sqrt{x}-2\sqrt{x}\right]_1^2=\frac{2}{3}(2)\sqrt{2}-2\sqrt{2}-\left(\frac{2}{3}(1)\sqrt{1}-2\sqrt{1}\right)\)

\(\displaystyle=\frac{4}{3}\sqrt{2}-2\sqrt{2}-\left(\frac{2}{3}-2\right)=-\frac{2}{3}\sqrt{2}-\left(-\frac{4}{3}\right)\)

\(\displaystyle=\frac{4}{3}-\frac{2}{3}\sqrt{2}\)

다음 적분을 구하라.

$$\int_0^1(1-x)^2dx$$

풀이

\(\displaystyle\int_0^1(1-x)^2dx=\left[-\frac{1}{3}(1-x)^3\right]_0^1\)

\(\displaystyle=-\frac{1}{3}(1-1)^3-\left(-\frac{1}{3}(1-0)^3\right)=\frac{1}{3}\)