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:

$intlimits^pi_0 {sinx} , dx=-cosx bigg|^pi_0=-cospi+cos0=-(-1)+1=2\ \ \intlimits^{2pi}_frac{2pi}{3}{cos(0,25x)} , dx=frac{1}{0,25}sin(0,25x)bigg |^{2pi}_frac{2pi}{3}=4sinfrac{x}{4}bigg|^{2pi}_frac{2pi}{3}=4sinfrac{pi}{2} -4sinfrac{pi}{6}=\ \ =4cdot1-4cdotfrac{1}{2}=4-=2$

$intlimits^frac{pi}{2}_{frac{pi}{4}}{sin2x} , dx=-frac{1}{2}cos2xbigg|^bigg{frac{pi}{2}} _bigg{{frac{pi}{4} }}=-frac{1}{2}cospi+frac{1}{2}cosfrac{pi}{2}=-frac{1}{2}cdot(-1)+frac{1}{2}cdot0=frac{1}{2}\ \\intlimits^4_0 {(x+frac{1}{sqrt{x} }) } , dx=intlimits^4_0 {(x+x^{-frac{1}{2}}) } , dx=(frac{x^2}{2}+{frac{x^{-frac{1}{2}+1}}{-frac{1}{2}+1}})bigg|^4_0=(frac{x^2}{2} +2sqrt{x})bigg|^4_0=\ \= frac{4^2}{2} +2sqrt{4}-frac{0^2}{2} -sqrt{0}=8+4=12$