Question

Решите
пж срочно надо
желательно во вложении

Answer 1

$1)\; \; \int \limits _{-1}^2\, 2\, dx=2x|_{-1}^2=2(2+1)=6\\\\2)\; \; \int \limits _1^3(x^2-2x)dx=(\frac{x^3}{3}-x^2)|_1^3=\frac{27}{3}-9-(\frac{1}{3}-1)=\frac{2}{3}\\\\3)\; \; \int \limits _1^8\, \sqrt[3]{x}dx=\frac{3x^{\frac{4}{3}}}{4}|_1^8=\frac{3}{4}(8^{\frac{4}{3}}-1)=\frac{3}{4}(16-1)=\frac{45}{4}\\\\4)\; \; \int \limits _0^{\frac{\pi}{2}}\, sinx\, dx=-cosx|_0^{\frac{\pi}{2}}-(cos\frac{\pi}{2}-cos0)=-(0-1)=1$

$5)\; \; \int \limits _{-1}^1(6x^3-5x)dx=(6\cdot \frac{x^4}{4}-5\cdot \frac{x^2}{2})|_{-1}^1=\frac{3}{2}-\frac{5}{2}-(\frac{3}{2}-\frac{5}{2})=0\\\\6)\; \; \int \limits _1^8\, 4\sqrt[3]{x}(1-\frac{4}{x})dx=4\cdot \int \limits _1^8\, (x^{\frac{1}{3}}-4\cdot x^{-\frac{2}{3}})dx=\\\\=4(\frac{3x^{\frac{4}{3}}}{4}-4\cdot \frac{x^{\frac{1}{3}}}{1/3})|_1^8=4(\frac{3}{4}\cdot 16-12\cdot 2)-4(\frac{3}{4}}-12)=-3$

$7)\; \; \int \limits _{2}^6\sqrt{2x-3}dx=\frac{1}{2}\cdot \frac{2(2x-3)^{\frac{3}{2}}}{3}|_2^6=\frac{1}{3}(9^{3/2}-1)=\frac{1}{3}(27-1)=\frac{26}{3}$

$8)\; \; \int \limits _0^{\frac{\pi}{4}}\frac{1}{2}cos(x+\frac{\pi}{4})dx=\frac{1}{2}sin(x+\frac{\pi}{4})|_0^{\frac{\pi}{4}}=\frac{1}{2}(sin\frac{\pi}{2}-sin\frac{\pi}{4})=\\\\=\frac{1}{2}(1-\frac{\sqrt2}{2})=\frac{2-\sqrt2}{4}\\\\9)\; \; \int \limits _1^3\; 3sin(3x-6)dx=-3\cdot \frac{1}{3}cos(3x-6)_1^3=\\\\=-(cos3-cos(-3))=-(cos3-cos3)=0$