Question

Вызначить интеграл
∫1/0(6/√3х+1 -х)dx

Answer 1

$\int\limits^1_0 { \frac{6}{ \sqrt{3x+1} } } \, dx=\int \frac{6}{ \sqrt{3x+1} }dx=6\int \frac{1}{ \sqrt{3x+1} }dx=6\int \frac{1}{3t}dt=6* \frac{1}{3}*\int \frac{1}{ \sqrt{t} }dt= \\ \\ \\ = 2*2 \sqrt{t}=\ \textgreater \ 2*2 \sqrt{3x+1}=4 \sqrt{3x+1} \\ \\ 4 \sqrt{3x+1} \large \mid^1_0=4 \sqrt{3*1+1} -4 \sqrt{3*0+1}=4 \sqrt{4}-4 \sqrt{1} =8-4=\boxed4$