Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn $I=\int\limits_{9}^{16}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}\text{d}x}=6$ và $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\cos x+1 \right)\sin x\text{d}x}=3$. Tính tích phân $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}$.
A. $I=-2$.
B. $I=6$.
C. $I=9$.
D. $I=2$.
A. $I=-2$.
B. $I=6$.
C. $I=9$.
D. $I=2$.
Xét $I=\int\limits_{9}^{16}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}\text{d}x}=6$, đặt $\sqrt{x}=t\Rightarrow \dfrac{\text{d}x}{2\sqrt{x}}=\text{d}t$
Đổi cận: $x=9\Rightarrow t=3$ ; $x=16\Rightarrow t=4$
$I=2\int\limits_{3}^{4}{f\left( t \right)\text{d}t}=6$ $\Rightarrow \int\limits_{3}^{4}{f\left( t \right)\text{d}t}=\dfrac{6}{2}=3$.
$J=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\cos x+1 \right)\sin x\text{d}x}=3$, đặt $2\cos x+1=u\Rightarrow -2\sin x\text{d}x=\text{d}u$
Đổi cận: $x=0\Rightarrow u=3$ ; $x=\dfrac{\pi }{2}\Rightarrow u=1$.
$J=-\dfrac{1}{2}\int\limits_{3}^{1}{f\left( u \right)\text{d}u}=3\Rightarrow \int\limits_{1}^{3}{f\left( u \right)du}=6$.
Vậy $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=\int\limits_{1}^{3}{f\left( x \right)\text{d}x}+\int\limits_{3}^{4}{f\left( x \right)\text{d}x}=6+3=9$.
Đổi cận: $x=9\Rightarrow t=3$ ; $x=16\Rightarrow t=4$
$I=2\int\limits_{3}^{4}{f\left( t \right)\text{d}t}=6$ $\Rightarrow \int\limits_{3}^{4}{f\left( t \right)\text{d}t}=\dfrac{6}{2}=3$.
$J=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\cos x+1 \right)\sin x\text{d}x}=3$, đặt $2\cos x+1=u\Rightarrow -2\sin x\text{d}x=\text{d}u$
Đổi cận: $x=0\Rightarrow u=3$ ; $x=\dfrac{\pi }{2}\Rightarrow u=1$.
$J=-\dfrac{1}{2}\int\limits_{3}^{1}{f\left( u \right)\text{d}u}=3\Rightarrow \int\limits_{1}^{3}{f\left( u \right)du}=6$.
Vậy $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=\int\limits_{1}^{3}{f\left( x \right)\text{d}x}+\int\limits_{3}^{4}{f\left( x \right)\text{d}x}=6+3=9$.
Đáp án C.