Cho hàm số $f\left(x \right)=\left\{ \begin{aligned} & {{x}^{2}}-5x+3\text{ khi }x\ge 7 \\ & 2x+3\text{ khi }x<7 \\ \end{aligned}...

Xét tích phân $I=\int\limits_{0}^{\ln 4}{f\left( 2{{e}^{x}}+3 \right){{e}^{x}}dx}.$
Đặt $t=2{{e}^{x}}+3\Rightarrow dt=2{{e}^{x}}dx$ hay ${{e}^{x}}dx=\dfrac{1}{2}dt.$
Đổi cận: $x=0\Rightarrow t=5;x=\ln 4\Rightarrow t=11.$
Khi đó:
$I=\dfrac{1}{2}\int\limits_{5}^{11}{f\left( t \right)dt}=\dfrac{1}{2}\int\limits_{5}^{11}{f\left( x \right)dx}=\dfrac{1}{2}\left( \int\limits_{5}^{7}{f\left( x \right)dx}+\int\limits_{7}^{11}{f\left( x \right)dx} \right)=\dfrac{1}{2}\left[ \int\limits_{5}^{7}{\left( 2x+3 \right)dx}+\int\limits_{7}^{11}{\left( {{x}^{2}}-5x+3 \right)dx} \right]$
$=\dfrac{1}{2}\left[ \left( {{x}^{2}}+3x \right)\left| \begin{aligned}
& 7 \\
& 5 \\
\end{aligned} \right.+\left( \dfrac{{{x}^{3}}}{3}-\dfrac{5{{x}^{2}}}{2}+3x \right)\left| \begin{aligned}
& 11 \\
& 7 \\
\end{aligned} \right. \right]=\dfrac{1}{2}\left( 30+\dfrac{484}{3} \right)=\dfrac{287}{3}.$
Vậy $\int\limits_{0}^{\ln 4}{f\left( 2{{e}^{x}}+3 \right){{e}^{x}}dx}=\dfrac{287}{3}.$