integrate x cos(x^2) dx, x=0..sqrt(pi)(b) $u=\sqrt{x}$로 치환하면 $du=\dfrac{1}{2\sqrt{x}}\,dx$이고, $\displaystyle\int_1^4\dfrac{e^{\sqrt{x}}}{\sqrt{x}}\,dx=\int_1^2\dfrac{e^u}\,(2du)=2\Bigl[e^u\Bigr]_1^2=2e(e-1)$
integrate e^(sqrt(x))/sqrt(x) dx, x=1..4(c) $\dfrac{x^2\sin x}{1+x^6}$는 기함수이므로 $\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{x^2\sin x}{1+x^6}\,dx=0$
integrate (x^2 sin(x))/(1+x^6) dx, x=-pi/2..pi/2
integrate (c^2-x^2)-(x^2-c^2) dx, x=-c..c
integrate (pi)(1-y^2)^2-(pi)(1-cbrt(y))^2 dy, y=0..1(셸 적분) $V=\displaystyle\int_0^1 2\pi(1-x)(\sqrt{x}-x^3)\,dx=2\pi\int_0^1(x^{1/2}-x^{3/2}-x^3+x^4)\,dx=2\pi\Bigl[\tfrac{2}{3}x^{3/2}-\tfrac{2}{5}x^{5/2}-\tfrac{1}{4}x^4+\tfrac{1}{5}x^5\Bigr]_0^1=2\pi(\tfrac{2}{3}-\tfrac{2}{5}-\tfrac{1}{4}+\tfrac{1}{5})=\dfrac{13}{30}\pi$
integrate 2(pi)(1-x)(sqrt(x)-x^3) dx, x=0..1(b) (디스크 적분) $A(x)=\pi(1-x^3)^2-\pi(1-\sqrt{x})^2$이므로 회전체의 부피는 $V=\displaystyle\int_0^1 A(x)\,dx=\pi\int_0^1(-2x^3+x^6+2x^{1/2}-x)\,dx=\pi\Bigl[-\tfrac{1}{2}x^4+\tfrac{1}{7}x^7+\tfrac{4}{3}x^{3/2}-\tfrac{1}{2}x^2\Bigr]_0^1=\pi(-\tfrac{1}{2}+\tfrac{1}{7}+\tfrac{4}{3}-\tfrac{1}{2})=\dfrac{10}{21}\pi$
integrate (pi)(1-x^3)^2-(pi)(1-sqrt(x))^2 dx, x=0..1(셸 적분) $V=\displaystyle\int_0^1 2\pi(1-y)(\sqrt[3]{y}-y^2)\,dy=2\pi\int_0^1(y^{1/3}-y^{4/3}-y^2+y^3)\,dy=2\pi\Bigl[\tfrac{3}{4}y^{4/3}-\tfrac{3}{7}x^{7/3}-\tfrac{1}{3}y^3+\tfrac{1}{4}y^4\Bigr]_0^1=2\pi(\tfrac{3}{4}-\tfrac{3}{7}-\tfrac{1}{3}+\tfrac{1}{4})=\dfrac{10}{21}\pi$
integrate 2(pi)(1-y)(cbrt(y)-y^2) dy, y=0..1
arc length of ln(cos(x)) from x=0..pi/3