Question

вычислите пожалуйста подробно определенный интеграл

Answer 1

$1) \int\limits^8_1 {\sqrt[3]{x^2} } \, dx =\int\limits^8_1 {x^{\frac{2}{3} } } \, dx=\frac{3x\sqrt[3]{x^2} }{5}\bigg|_1^8= \Big(\frac{3\cdot8\sqrt[3]{2^6} }{5}\Big)-\Big(\frac{3\cdot1\sqrt[3]{1^2} }{5}\Big)=\frac{3\cdot8\cdot4-3}{5}=\\=\frac{93}{5}=\boxed{18\frac{3}{5} }$

$2)\int\limits^3_1 {e^{2x}} \, dx = \frac{e^{2x}}{2}\bigg|_1^3=\frac{e^6-e^2}{2}=\boxed{\frac{e^2(e^4-1)}{2}}$

$3)\int\limits^1_0 {\frac{dx}{x+2}=\ln|x+2|\bigg|_0^1=\ln3-\ln2=\boxed{\ln1,5}$

$4)\int\limits^{\frac{\pi}{4}}_0 {\cos{2x}} \, dx =\frac{1}{2}\sin2x\bigg|_0^{\frac{\pi}{4}}=\frac{1}{2}\big(\sin{\frac{\pi}{2}-\sin0}\big)=\boxed{\frac{1}{2}}$

$5)\int\limits^1_0 {\frac{dx}{\cos^2{\frac{x}{2}}}} =2\tan{\frac{x}{2}}\bigg|_0^1=2\tan{\frac{1}{2}}-2\tan0=\boxed{2\tan{\frac{1}{2}}}$

$6)\int\limits^{\sqrt3}_{\sqrt2} {\frac{dx}{\sqrt{4-x^2}} }= \int\limits^{\sqrt3}_{\sqrt2} {\frac{dx}{\sqrt{2^2-x^2}} }=\arcsin{\frac{x}{2}}\bigg|_{\sqrt{2}}^{\sqrt{3}}=\arcsin{\frac{\sqrt{3}}{2}}-\arcsin{\frac{\sqrt{2}}{2}}=\frac{\pi}{3} -\frac{\pi}{4}=\boxed{\frac{\pi}{12} }$

$7)\int\limits^{2\sqrt2}_{\sqrt5} {\frac{xdx}{\sqrt{3x^2+1}} }=\left[t=3x^2+1;dt=6xdx\atop{t_1=25;t_2=16}\right]=\frac{1}{6}\int\limits^{25}_{16} {\frac{dt}{\sqrt{t}} }=\frac{\sqrt{t} }{3}\bigg|_{16}^{25}=\frac{5}{3}-\frac{4}{3}=\boxed{\frac{1}{3}}$

$8)\int\limits^1_0 {\arcsin{x}} \, dx =\left[u=\arcsin{x};du=\frac{dx}{\sqrt{1-x^2}}\atop dv=dx;v=x\right]=x\arcsin{x}\bigg|_0^1-\int\limits^1_0\frac{xdx}{\sqrt{1-x^2}}=(*)$

$\int\limits^1_0\frac{xdx}{\sqrt{1-x^2}}=\int\limits^1_0\frac{d(x^2)}{2\sqrt{1-x^2}}=\lim\limits_{a \to 1-0}\bigg(\int\limits^a_0\frac{d(x^2)}{2\sqrt{1-x^2}}\bigg)=\lim\limits_{a \to 1-0}\bigg(-\sqrt{1-x^2}\bigg|_0^a}\bigg)=\\=\lim\limits_{a \to 1-0}\bigg(-\sqrt{1-x^2}\bigg|_0^a}\bigg)=\lim\limits_{a \to 1-0}\bigg(-\sqrt{1-a^2}+1\bigg)=1$

$(*)=\arcsin1-\arcsin0-1=\boxed{\frac{\pi}{2}-1}$

$9)\int\limits^{\frac{5\pi}{4}}_\frac{\pi}{4} {{\sin7x\cdot\cos3x}\, dx=\frac{1}{2}\int\limits^{\frac{5\pi}{4}}_\frac{\pi}{4} {{\sin4x+\sin10x}\, dx=\frac{-\cos4x}{8}\bigg|_\frac{\pi}{4}^\frac{5\pi}{4}-\frac{cos10x}{20}\bigg|_\frac{\pi}{4}^\frac{5\pi}{4}=$

$=\frac{1}{8}-\frac{1}{8}-\bigg(-\frac{\cos{\frac{25\pi}{2}}}{20}+\frac{\cos{\frac{5\pi}{2}}}{20}\bigg)=\frac{\cos{(12\pi+\frac{\pi}{2}})}{20}+\frac{\cos{(2\pi+\frac{\pi}{2}})}{20}=\boxed0$

$10)\int\limits^2_1 {x\sin{x}} \, dx =\left[u=x;du=dx\atop dv=\sin{x}dx;v=-\cos{x}\right]=-x\cos{x}\bigg|_1^2+\int\limits^2_1{\cos{x}}\,dx=-2\cos2++\cos1+\sin{x}\bigg|_1^2=\boxed{-2\cos2+\cos{1}+\sin2-\sin1}$