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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& 1-{{x}^{2}}\text{khi} x\le 3 \\
& 7-5x \text{khi} x>3 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{\ln 2}{f\left( 3{{\text{e}}^{x}}-1 \right){{\text{e}}^{x}} \text{d}x}$.
A. $\dfrac{13}{15}$.
B. $-\dfrac{94}{9}$.
C. $-\dfrac{102}{33}$.
D. $\dfrac{25}{9}$.
& 1-{{x}^{2}}\text{khi} x\le 3 \\
& 7-5x \text{khi} x>3 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{\ln 2}{f\left( 3{{\text{e}}^{x}}-1 \right){{\text{e}}^{x}} \text{d}x}$.
A. $\dfrac{13}{15}$.
B. $-\dfrac{94}{9}$.
C. $-\dfrac{102}{33}$.
D. $\dfrac{25}{9}$.
Đặt $u=3{{\text{e}}^{x}}-1\Rightarrow \dfrac{1}{3}\text{d}u={{\text{e}}^{x}}\text{d}x$.
Đổi cận $x=0\Rightarrow u=2$ ; $x=\ln 2\Rightarrow u=5$.
Ta có $\int\limits_{0}^{\ln 2}{f\left( 3{{\text{e}}^{x}}-1 \right){{\text{e}}^{x}} \text{d}x}=\dfrac{1}{3}\int\limits_{2}^{5}{f\left( u \right)\text{d}u}=\dfrac{1}{3}\int\limits_{2}^{3}{\left( 1-{{u}^{2}} \right)\text{d}u}+\dfrac{1}{3}\int\limits_{3}^{5}{\left( 7-5u \right)\text{d}u}=-\dfrac{94}{9}$.
Đổi cận $x=0\Rightarrow u=2$ ; $x=\ln 2\Rightarrow u=5$.
Ta có $\int\limits_{0}^{\ln 2}{f\left( 3{{\text{e}}^{x}}-1 \right){{\text{e}}^{x}} \text{d}x}=\dfrac{1}{3}\int\limits_{2}^{5}{f\left( u \right)\text{d}u}=\dfrac{1}{3}\int\limits_{2}^{3}{\left( 1-{{u}^{2}} \right)\text{d}u}+\dfrac{1}{3}\int\limits_{3}^{5}{\left( 7-5u \right)\text{d}u}=-\dfrac{94}{9}$.
Đáp án B.