[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) \(\int_0^{\pi/2}\sin^3\)

--- under construction ---