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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
\(\displaystyle\int_0^1(1-x)^2dx=\left[-\frac{1}{3}(1-x)^3\right]_0^1\)
\(\displaystyle=-\frac{1}{3}(1-1)^3-\left(-\frac{1}{3}(1-0)^3\right)=\frac{1}{3}\)
수직선 위를 움직이고 있는 어떤 입자는 시간 \(t\)에서의 가속도가 \(a(t)=6t+4\), 초기 속도는 \(v(0)=-6\), 초기 위치는 \(s(0)=3\)이라고 한다. 이때 운동방정식 \(s(t)\)를 구하라.
풀이
\(v'(t)=a(t)\)
\(\displaystyle v(t)=\int{6t+4}dt=\frac{6}{2}t^2+4t+C=3t^2+4t+C\)
\(v(0)=-6=3(0)^2+4(0)+C=C\)
\(v(t)=3t^2+4t-6\)
\(s'(t)=v(t)\)
\(\displaystyle s(t)=\int{3t^2+4t-6}dt=\frac{3}{3}t^3+\frac{4}{2}t^2-6t+C=t^3+2t^2-6t+C\)
\(s(0)=3=(0)^3+2(0)^2-6(0)+C=C\)
\(s(t)=t^3+2t^2-6t+3\)
\(\displaystyle f(x)=\frac{1}{4}\sqrt{x}-\frac{1}{x^2}\)의 부정적분 \(F(x)\)가 \(\displaystyle F(4)=\frac{5}{4}\)일 때, \(F(x)\)를 구하라.
풀이
\(\displaystyle F(x)=\int\left(\frac{1}{4}\sqrt{x}-\frac{1}{x^2}\right)dx=\int\left(\frac{1}{4}x^{\frac{1}{2}}-x^{-2}\right)dx\)
\(\displaystyle=\frac{1}{4}\cdot\frac{2}{3}x^{\frac{3}{2}}-(-x^{-1})+C=\frac{1}{6}\sqrt{x^3}+x^{-1}+C\)
\(\displaystyle=\frac{1}{6}x\sqrt{x}+\frac{1}{x}+C\)
\(\displaystyle F(4)=\frac{5}{4}=\frac{1}{6}4\sqrt{4}+\frac{1}{4}+C\)
\(\displaystyle C=\frac{5}{4}-\frac{1}{6}4\cdot2-\frac{1}{4}=1-\frac{4}{3}=-\frac{1}{3}\)
\(\displaystyle F(x)=\frac{1}{6}x\sqrt{x}+\frac{1}{x}-\frac{1}{3}\)
다음 부정적분을 구하라.
$$\int\sqrt{2x-1}dx$$
풀이
\(\displaystyle\int\sqrt{2x-1}dx=\int(2x-1)^{\frac{1}{2}}dx\)
\(\displaystyle=\frac{1}{2}\cdot\frac{2}{3}(2x-1)^{\frac{3}{2}}+C=\frac{1}{3}(2x-1)^{\frac{3}{2}}+C\)
다음 부정적분을 구하라.
$$\int\frac{1}{(2x+3)^4}dx$$
풀이
\(\displaystyle\int\frac{1}{(2x+3)^4}dx=\int(2x+3)^{-4}dx\)
\(\displaystyle=\frac{1}{2}\cdot\frac{1}{-3}(2x+3)^{-3}+C=-\frac{1}{6(2x+3)^3}+C\)
다음 부정적분을 구하라.
$$\int(4x+1)^3dx$$
풀이
\(\displaystyle\int(4x+1)^3dx=\frac{1}{4}\cdot\frac{1}{4}(4x+1)^4+C\)
\(\displaystyle=\frac{1}{16}(4x+1)^4+C\)