Задача 72022 Вычислить 5 интегралов [b]n=8[/b]...
Условие
Решение
[m] ∫ ^{8}_{1}\frac{5x^2+7x}{x}dx=[/m]
Делим почленно каждое слагаемое числителя на знаменатель
[m] =∫ ^{8}_{1}(\frac{5x^2}{x}+\frac{5x^2+7x}{x})dx=∫ ^{8}_{1}(5x+7)dx=(5\frac{x^2}{2}+7x)|^{8}_{1}=(5\frac{8^2}{2}+7\cdot 8)-(5\frac{1^2}{2}+7\cdot 1)=[/m]
2.
[m] ∫ ^{8}_{0}\frac{dx}{8^2+x^2}=\frac{1}{8}(arctg\frac{x}{8})|^{8}_{0}=\frac{1}{8}arctg\frac{8}{8}-\frac{1}{8}arctg0=\frac{1}{8}arctg1=\frac{1}{8}\cdot \frac{π}{4}=\frac{π}{32}[/m]
3.
[m]∫^{8}_{0}\frac{dx}{\sqrt{8x}}dx=\frac{1}{\sqrt{8}}∫ ^{8}_{0}\frac{1}{\sqrt{x}}dx=\frac{1}{\sqrt{8}}\cdot 2(\sqrt{x})| ^{8}_{0}=\frac{1}{\sqrt{8}}\cdot 2\sqrt{8}=2[/m]
4.
[m] ∫ ^{\frac{π}{3}}_{0}8cos3xdx=\frac{8}{3}∫ ^{\frac{π}{3}}_{0}cos3x d(3x)=\frac{8}{3}\cdot (sin3x)|^{\frac{π}{3}}_{0}=\frac{8}{3}\cdot (sinπ-sin0)=0[/m]
5.
[m] ∫ ^{\frac{π}{8}}_{0}x\cdot cos8xdx=[/m]
интегрирование по частям
u=x
dv=cos8xdx
du=dx
v= ∫ cos8xdx=(1/8) ∫ cos8x d(8x)=(1/8)sin8x
[m] =\frac{xsin8x}{8}|^{\frac{π}{8}}_{0}-∫ ^{\frac{π}{8}}_{0}\frac{1}{8}sin8xdx=\frac{\frac{π}{8}sin8\frac{π}{8}-0 sin 0}{8}+\frac{1}{64}(cos8x)|^{\frac{π}{8}}_{0}=\frac{1}{64}(cosπ-cos0)=\frac{1}{64}(-1-1)=-\frac{1}{32}[/m]