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Câu hỏi: Cho hàm số $y=f(x)=\left\{\begin{array}{l}x^2+3 \text { khi } x \geq 1 \\ 5-x \text { khi } x<1\end{array}\right.$.
Tính $I=2 \int_0^{\dfrac{\pi}{2}} f(\sin x) \cos x d x+3 \int_0^1 f(3-2 x) d x$
A. $I=\dfrac{71}{6}$.
B. $I=32$.
C. $I=\dfrac{32}{2}$.
D. $I=31$.
Tính $I=2 \int_0^{\dfrac{\pi}{2}} f(\sin x) \cos x d x+3 \int_0^1 f(3-2 x) d x$
A. $I=\dfrac{71}{6}$.
B. $I=32$.
C. $I=\dfrac{32}{2}$.
D. $I=31$.
+ Tính $\int_0^{\dfrac{\pi}{2}} f(\sin x) \cos x d x$. Đặt $\sin x=t \Rightarrow \cos x d x=d t$. Đổi cận $\left\{\begin{array}{l}x=0 \Rightarrow t=0 \\ x=\dfrac{\pi}{2} \Rightarrow t=1\end{array}\right.$
Do đó $\int_0^{\dfrac{\pi}{2}} f(\sin x) \cos x d x=\int_0^1 f(t) d t=\int_0^1(5-t) d t=\left(5 t-\dfrac{t^2}{2}\right) \mid \quad 1$.
+ Tính $\int_0^1 f(3-2 x) d x$. Đặt $t=3-2 x \Rightarrow d t=-2 d x \Rightarrow d x=\dfrac{-d t}{2}$
Đổi cận $\left\{\begin{array}{l}x=0 \Rightarrow t=3 \\ x=1 \Rightarrow t=1\end{array}\right.$
Do đó $\int_0^1 f(3-2 x) d x=\int_3^1 f(t) \cdot \dfrac{-d t}{2}=\dfrac{1}{2} \int_1^3 f(t) d t=\dfrac{1}{2} \int_1^3\left(x^2+3\right) d t=\dfrac{1}{2}\left(\dfrac{x^3}{3}+3 x\right) \mid$
3
Vậy $I=2 \cdot \dfrac{9}{2}+3 \cdot \dfrac{22}{3}=31$
Do đó $\int_0^{\dfrac{\pi}{2}} f(\sin x) \cos x d x=\int_0^1 f(t) d t=\int_0^1(5-t) d t=\left(5 t-\dfrac{t^2}{2}\right) \mid \quad 1$.
+ Tính $\int_0^1 f(3-2 x) d x$. Đặt $t=3-2 x \Rightarrow d t=-2 d x \Rightarrow d x=\dfrac{-d t}{2}$
Đổi cận $\left\{\begin{array}{l}x=0 \Rightarrow t=3 \\ x=1 \Rightarrow t=1\end{array}\right.$
Do đó $\int_0^1 f(3-2 x) d x=\int_3^1 f(t) \cdot \dfrac{-d t}{2}=\dfrac{1}{2} \int_1^3 f(t) d t=\dfrac{1}{2} \int_1^3\left(x^2+3\right) d t=\dfrac{1}{2}\left(\dfrac{x^3}{3}+3 x\right) \mid$
3
Vậy $I=2 \cdot \dfrac{9}{2}+3 \cdot \dfrac{22}{3}=31$
Đáp án D.