Question

cho f(x) và g(x) có đạo hàm trên [0;6] thỏa f(1) + g(1)=4; g(x)=-xf'(x); f(x)=-xg'(x). tính \(\int_1^4\left(f\left(x\right)+g\left(x\right)\right)dx\)

mọi người giúp em với ạ.

Answer 1

\(f\left(x\right)+g\left(x\right)=-x\left[f'\left(x\right)+g'\left(x\right)\right]\)

Đặt \(h\left(x\right)=f\left(x\right)+g\left(x\right)\Rightarrow\left\{{}\begin{matrix}h\left(1\right)=4\\h\left(x\right)=-x.h'\left(x\right)\end{matrix}\right.\)

\(\Rightarrow\frac{h'\left(x\right)}{h\left(x\right)}=-\frac{1}{x}\Rightarrow\int\frac{h'\left(x\right)}{h\left(x\right)}dx=-\int\frac{dx}{x}=-lnx\)

\(\Rightarrow ln\left[h\left(x\right)\right]=ln\left(\frac{1}{x}\right)+C\)

Thay \(x=1\Rightarrow C=ln4\Rightarrow ln\left[h\left(x\right)\right]=ln\left(\frac{1}{x}\right)+ln4=ln\left(\frac{4}{x}\right)\)

\(\Rightarrow h\left(x\right)=\frac{4}{x}\)

\(\Rightarrow I=\int\limits^4_1h\left(x\right)dx=\int\limits^4_1\frac{4}{x}dx=...\)