Question

СРОЧНО 100 БАЛОВ! ВЫЧИСЛЕНИЕ ИНТЕГРАЛА! ОБЧИСЛЕННЯ ІНТЕГРАЛА!
$1. \int\limits^5_4 {(4-x)^{3} } \, dx \\ 2.\int\limits^3_0 {\sqrt[3]{3x-1} } \, dx$

Answer 1

$\displaystyle \int\limits^5_4 {(4-x)^3} \, dx=\bigg|\frac{4-x=t; -dx=dt }{\int {-t^3dt}=-\frac{t^4}{4} } \bigg|=-\frac{(4-x)^4}{4}\bigg|^5_4=\\\\=-(\frac{(4-5)^4}{4}-\frac{(4-4)^4}{4})=-(\frac{1}{4}-0)=-\frac{1}{4}$

$\displaystyle \int\limits^3_0 {(3x-1)^{1/3}} \, dx =\\\\\bigg|3x-1=t; 3dx=dt; dx=\frac{1}{3}dt\\\\\int {\frac{1}{3} t^{1/3}} \, dt =\frac{1}{3}*\frac{t^{4/3}}{4/3}=\frac{1}{4}*t^{4/3}=\frac{1}{4}*\sqrt[3]{t^4}\bigg|\\\\=\frac{1}{4}(\sqrt[3]{(3x-1)^4})\bigg|^3_0=\frac{1}{4}(\sqrt[3]{8^4}-\sqrt[3]{(-1)^4})=\frac{1}{4}(2^4-1)=3.75$