[연습문제] 정적분

1. 다음 정적분의 값을 구하여라.

(1) \(\int_{-1/2}^{1/2}(2y+1)^7dy={1\over16}[(2y+1)^8]_{-1/2}^{1/2}=16\)

(2) \(\int_0^{\pi/2}\sqrt{\sin{x}}\cos{x}dx={2\over3}[\sin^{3/2}x]_0^{\pi/2}={2\over3}\)

(3) \(\int_0^1\frac{{\rm Tan^{-1}}x}{1+x^2}dx={1\over2}[({\rm Tan}^{-1})^2]_0^1={\pi^2\over32}\)

(4) \(\int_{-1}^1|2x-1|dx=\int_{-1}^{1/2}(-2x+1)dx+\int_{1/2}^1(2x-1)dx=[-x^2+x]_{-1}^{1/2}+[x^2-x]_{1/2}^1\)

                                    \(={5\over2}\)

2. 다음 정적분의 값을 구하여라.

(1) \(\int_1^4\frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int_1^2e^tdt=2[e^t]_1^2=9.3415\)

(2) \(\int_0^1(e^x+x^e)dx=\left[e^x+\frac{x^{e+1}}{e+1}\right]_0^1=1.9872\)

(3) \(\int_0^{\pi/3}\frac{\sec\theta\tan\theta}{\sqrt{e^{\sec\theta}}}d\theta=\int_1^2\frac{dx}{\sqrt{e^x}}=-2\left[{1\over e}-{1\over\sqrt{e}}\right]_1^2=0.4773\)

(4) \(\int_0^1(e^x+e^{-x})^2dx=\int_0^1(e^{2x}+2+e^{-2x})dx=\left[{e^{2x}\over2}+2x-{e^{-2x}\over2}\right]_0^1=5.6269\)

(5) \(\int_{-1}^0\frac{d\theta}{1+e^\theta}=\int_{1+1/e}^2\frac{dx}{x(x-1)}=\left[\ln\left|x-1\over x\right|\right]_{1+1/e}^2=0.6201\)

(6) \(\int_0^{\pi/2}\sin^3\theta d\theta=\int_0^{\pi/2}(1-\cos^2\theta)\sin\theta d\theta=\left[-\cos\theta+{\cos^3\theta\over3}\right]_0^{\pi/2}={2\over3}\)

(7) \(\int_{-\pi/4}^{\pi/4}\sec^6tdt=\int_{-\pi/4}^{\pi/4}(\tan^2t+1)\sec^2tdt=\int_{-1}^1(x^2+1)^2dx=\left[{x^5\over5}+{2x^3\over3}+x\right]_{-1}^1\)

                                  \(={56\over15}\)

(8) \(\int_0^\pi\frac{\cos^2x}{1-\sin{x}}dx=\int_0^\pi(1+\sin{x})dx=[x-\cos{x}]_0^\pi=\pi+2\)

3. 다음 정적분의 값을 구하여라.

(1) \(\int_3^5\sqrt{x^2-9}dx={1\over2}\left[x\sqrt{x^2-9}-9\ln(x+\sqrt{x^2-9})\right]_3^5=10-{9\over2}\ln3\)

적분공식 총정리 No. 18 참조

(2) \(\int_0^1\frac{dx}{2-x^2}=-\int_0^1\frac{dx}{x^2-2}=-{1\over2\sqrt{2}}\left[\ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|\right]_0^1={\sqrt{2}\over2}\ln(1+\sqrt{2})\)

적분공식 총정리 No. 11 참조

(3) \(\int_{1/2}^1\sqrt{3+4x-4x^2}dx=\int_{1/2}^1\sqrt{4-(1-2x)^2}dx\)

                                                    \(={1\over4}\left[(1-2x)\sqrt{3+4x-4x^2}+4{\rm Sin}^{-1}{1-2x\over2}\right]_{1/2}^1={\sqrt{3}\over2}+{\pi\over6}\)

적분공식 총정리 No. 17 참조

(4) \(\int_{\ln2}^{\ln3}\frac{dx}{e^x-e^{-x}}=\int_2^3\frac{dt}{t^2-1}={1\over2}\left[\ln\frac{t-1}{t+1}\right]_2^3={1\over2}\ln{3\over2}\)

적분공식 총정리 No. 17 참조

4. 다음 정적분의 값을 구하여라.

(1) \(\int_{-1}^1{\rm Cos}^{-1}xdx=[x{\rm Cos}^{-1}x]_{-1}^1+\int_{-1}^1\frac{x}{\sqrt{1-x^2}}dx=\pi+[\sqrt{1-x^2}]_{-1}^1=\pi\)

역함수의 미분법 참조

(2) \(\int_0^\sqrt{7}\frac{x^3}{\sqrt[3]{8-x^2}}dx={3\over2}\int_2^1(t^4-8t)dt={3\over2}\left[{t^5\over5}-4t^2\right]_2^1={87\over10}\)

(3) \(\int_0^{\pi/3}\sin3\theta\cos\theta d\theta=\int_0^{\pi/3}(-4\sin^3\theta+3\sin\theta)\cos\theta d\theta=\left[-\sin^4\theta+{3\over2}\sin^2\theta\right]_0^{\pi/3}={9\over16}\)

삼각함수 항등식 참조

(4) \(\int_0^1\frac{xe^x}{(1+x)^2}dx=\left[-\frac{xe^x}{1+x}\right]_0^1+\int_0^1e^xdx=-{e\over2}+[e^x]_0^1={e\over2}-1\)

5. 다음 정적분의 값을 구하여라.

(1) \(\int_1^3x\sqrt[3]{x^2-1}dx={3\over8}[(x^2-1)^{4/3}]_1^3=6\)

(2) \(\int_4^9\frac{dx}{\sqrt{x}-1}=2\int_2^3\left({1\over t-1}+1\right)dt=2[\ln(t-1)+t]_2^3=2\ln2+2\)

(3) \(\int_{-1}^0x^2\sqrt{x+1}dx=2\int_0^1(t^6-2t^4+t^2)dt=2\left[{t^7\over7}-{2t^5\over4}+{t^3\over3}\right]_0^1={16\over105}\)

(4) \(\int_0^7\frac{dy}{1+\sqrt[3]{y+1}}=3\int_1^2\left(x-1+{1\over x+1}\right)dx=3\left[{x^2\over2}-x+\ln(x+1)\right]_1^2={3\over2}+3\ln{3\over2}\)

6. 다음 정적분의 값을 구하여라.

(1) \(\int_5^9\frac{ds}{s\sqrt{x^2+144}}=\int_{13}^{15}\frac{dx}{x^2-144}={1\over24}\left[\ln\frac{x-12}{x+12}\right]_{13}^{15}={1\over12}\ln{5\over3}\)

적분공식 총정리 No. 11 참조

(2) \(\int_0^2x^2\sqrt{4-x^2}dx=4\int_0^{\pi/2}\sin^22\theta d\theta=2\int_0^{\pi/2}(1-\cos4\theta)d\theta=2\left[\theta-{\sin4\theta\over4}\right]_0^{\pi/2}=\pi\)

(3) \(\int_3^6\frac{dr}{r^3\sqrt{r^2-9}}=\int_0^{3\sqrt{3}}\frac{dx}{(x^2+9)^2}={1\over18}\left[\frac{x}{x^2+9}+{1\over3}{\rm Tan}^{-1}{x\over3}\right]_0^{3\sqrt{3}}=\frac{3\sqrt{3}+4\pi}{648}\)

유리함수의 적분 참조

(4) \(\int_1^\infty\frac{dx}{x^4\sqrt{x^2+3}}=\int_{\pi/6}^{\pi/2}\left(\frac{1-\sin^2t}{\sin^2t}\right)\cos{t}dt={1\over9}\left[{1\over\sin{t}}-{1\over3\sin^3t}\right]_{\pi/6}^{\pi/2}={4\over27}\)

7. 다음 정적분의 값을 계산하여라.

(1) \(\int_2^4\frac{x}{(x+1)(x+2)}dx=\int_2^4\left({2\over x+2}-{1\over x+1}\right)dx=\left[\ln\frac{(x+2)^2}{x+1}\right]_2^4=\ln{27\over20}\)

(2) \(\int_1^2\frac{5x^2-3x+18}{x(9-x^2)}dx=\int_1^2\left({2\over x}-{4\over x+3}-{3\over x-3}\right)dx=[2\ln{x}-4\ln(x+3)-3\ln|x-3|]_1^2\)

                                      \(=13\ln2-4\ln5\)

(3) \(\int_2^3\frac{4y}{(y-1)(y^2-1)}dy=\int_2^3\left\{{1\over y-1}+{2\over(y-1)^2}-{1\over y+1}\right\}dy=\left[\ln\frac{y-1}{y+1}-{2\over y-1}\right]_2^3=1+\ln{3\over2}\)

(4) \(\int_0^1\frac{dx}{1+x^3}={1\over3}\int_0^1\left[\frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2\left\{\left(x-{1\over2}\right)^2+{3\over4}\right\}}\right]dx\)

                       \(={1\over3}\left[\ln\frac{x+1}{\sqrt{x^2-x+1}}+\sqrt{3}{\rm Tan}^{-1}\frac{2x-1}{\sqrt{3}}\right]_0^1={\ln2\over3}+{\pi\over3\sqrt{3}}\)

(5) \(\int_0^1\frac{x^2+3x+1}{x^4+x^2+1}dx=\int_0^1\left\{\frac{2}{\left(x-{1\over2}\right)^2+{3\over4}}-\frac{1}{\left(x+{1\over2}\right)^2+{3\over4}}\right\}dx\)

                                   \(={2\over\sqrt{3}}\left[2{\rm Tan}^{-1}\frac{2x-1}{\sqrt{3}}-{\rm Tan}^{-1}\frac{2x+1}{\sqrt{3}}\right]_0^1={\pi\over\sqrt{3}}\)

이상 유리함수의 적분 참조

8. 다음 이상적분이 수렴하면 그 값을 구하여라.

(1) \(\int_1^\infty\frac{dx}{x^3}=-{1\over2}\left[{1\over x^2}\right]_1^\infty={1\over2}\)

(2) \(\int_{-\infty}^1e^xdx=[e^x]_{-\infty}^1=e\)

(3) \(\int_5^\infty\frac{dy}{\sqrt{y-1}}=2\left[\sqrt{y-1}\right]_5^\infty=\infty\)

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

(5) \(\int_0^{\pi/2}\tan\theta d\theta=-[\ln\cos\theta]_0^{\pi/2}=\infty\)

(6) \(\int_0^2\frac{dx}{(x-1)^{2/3}}=3\left[\sqrt[3]{x-1}\right]_0^2=6\)

(7) \(\int_1^{2.6}\frac{dy}{\sqrt{y^2-1}}=\left[\ln\left(y+\sqrt{y^2-1}\right)\right]_1^{2.6}=\ln5\)

(8) \(\int_2^\infty\frac{dx}{x\sqrt{x^2-4}}={1\over2}\int_0^{\pi/2}dt={1\over2}[t]_0^{\pi/2}={\pi\over4}\)

(9) \(\int_0^4\frac{dx}{x^2-2x-3}={1\over4}\int_0^4\left({1\over x-3}-{1\over x+1}\right)dx={1\over4}\left(\left[\ln\left|\frac{x-3}{x+1}\right|\right]_0^3+\left[\ln\left|\frac{x-3}{x+1}\right|\right]_3^4\right)=-\infty+\infty\)

9. 사다리꼴의 공식을 사용하여 다음 적분의 근사값을 구하고, 적분계산으로 그 결과를 검토하여라.

(1) \(\int_2^7\frac{dx}{x^2},\quad n=5\)          (2) \(\int_3^6\frac{dx}{\sqrt{x-2}},\quad n=6\)

<풀이>

(1) \(h=(b-a)/n=(7-2)/5=1\)

\(x_i\)	2	3	4	5	6	7
\(y_i\)	1/4	1/9	1/16	1/25	1/36	1/49

\(\int_2^7\fallingdotseq{h\over2}\{y_0+2(y_1+\cdots+y_4)+y_5\}={1\over2}\left\{{1\over4}+2\left({1\over9}+{1\over16}+{1\over25}+{1\over36}\right)+{1\over49}\right\}=0.377\)

(2) \(h=(b-a)/n=(6-3)/3=0.5\)

\(x_i\)	3.0	3.5	4.0	4.5	5.0	5.5	6.0
\(y_i\)	1.000	0.816	0.707	0.632	0.577	0.535	0.500

\(\int_3^6=\frac{dx}{\sqrt{x-2}}\fallingdotseq{h\over2}\{y_0+2(y_1+\cdots+y_5)+y_6\}\)

                        \(={0.5\over2}\{1.000+2(0.816+0.707+0.632+0.577+0.535)+0.500\}=2.009\)

10. Simpson의 공식을 사용하여 다음 적분의 근사값을 구하고, 적분계산으로 그 결과를 검토하여라.

(1) \(\int_2^8x^3dx,\quad n=2\)          (2) \(\int_0^4\frac{x}{\sqrt{x^2+9}}dx,\quad n=4\)

<풀이>

(1) \(h=(b-a)/(2n)=(8-2)/(2\times2)=1.5\)

\(x_i\)	2.0	3.5	5.0	6.5	8.0
\(y_i\)	8.000	42.875	125.000	274.625	512.000

\(\int_2^8x^3dx\approx{h\over3}\{y_0+4(y_1+y_3)+2y_2+y_4\}\)

                 \(={1.5\over3}\{8.000+4(42.875+274.625)+2(125.000)+512.000\}=1020.000\)

(2) \(h=(b-a)/(2n)=(4-0)/(2\times4)=0.5\)

\(x_i\)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
\(y_i\)	0.000	0.164	0.316	0.447	0.555	0.640	0.707	0.756	0.800

\(\int_0^4\frac{x}{\sqrt{x^2+9}}dx\approx{h\over3}\{y_0+4(y_1+y_3+y_5+y_7)+2(y_2+y_4+y_6)+y_8\}\)

                         \(={0.5\over3}\{0.000+4(0.164+0.447+0.640+0.759)+2(0.316+0.555\)

                         \(+0.707)+0.800=2.000\)