Question

Интеграл Алгебра 10-11 класс.

Answer 1

Правила:

$\int {sinx} \, dx =-cosx+c\\ \\\int {cosx} \, dx=sinx+c \\ \\ \int {cos(kx)} \, dx=\frac{1}{k}sin(kx)+c\\ \\ \int {sin(kx)} \, dx=-\frac{1}{k}cos(kx)+c\\ \\ \int {x^n} \, dx =\frac{x^{n+1}}{n+1} +c$

Решение:

$\int\limits^\pi_0 {sinx} \, dx=-cosx \bigg|^\pi_0=-cos\pi+cos0=-(-1)+1=2\\ \\ \\\int\limits^{2\pi}_\frac{2\pi}{3}{cos(0,25x)} \, dx=\frac{1}{0,25}sin(0,25x)\bigg |^{2\pi}_\frac{2\pi}{3}=4sin\frac{x}{4}\bigg|^{2\pi}_\frac{2\pi}{3}=4sin\frac{\pi}{2} -4sin\frac{\pi}{6}=\\ \\ =4\cdot1-4\cdot\frac{1}{2}=4-=2$

$\int\limits^\frac{\pi}{2}_{\frac{\pi}{4}}{sin2x} \, dx=-\frac{1}{2}cos2x\bigg|^\bigg{\frac{\pi}{2}} _\bigg{{\frac{\pi}{4} }}=-\frac{1}{2}cos\pi+\frac{1}{2}cos\frac{\pi}{2}=-\frac{1}{2}\cdot(-1)+\frac{1}{2}\cdot0=\frac{1}{2}\\ \\\\\int\limits^4_0 {(x+\frac{1}{\sqrt{x} }) } \, dx=\int\limits^4_0 {(x+x^{-\frac{1}{2}}) } \, dx=(\frac{x^2}{2}+{\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}})\bigg|^4_0=(\frac{x^2}{2} +2\sqrt{x})\bigg|^4_0=\\ \\= \frac{4^2}{2} +2\sqrt{4}-\frac{0^2}{2} -\sqrt{0}=8+4=12$