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Câu hỏi: Biết $\int_{-1}^{11} f(x) \mathrm{d} x=18$. Tính $I=\int_0^2 x\left[2+f\left(3 x^2-1\right)\right] \mathrm{d} x$.
A. $I=8$.
B. $I=10$.
C. $I=5$.
D. $I=7$.
A. $I=8$.
B. $I=10$.
C. $I=5$.
D. $I=7$.
Ta có $I=\int_0^2 x\left[2+f\left(3 x^2-1\right)\right] \mathrm{d} x=\int_0^2 2 x \mathrm{~d} x+\int_0^2 x \cdot f\left(3 x^2-1\right) \mathrm{d} x=I_1+I_2$.
+) $I_1=\int_0^2 2 x \mathrm{~d} x=\left.x^2\right|_0 ^2=4$.
+) $I_2=\int_0^2 x . f\left(3 x^2-1\right) \mathrm{d} x=\dfrac{1}{6} \int_0^2 f\left(3 x^2-1\right) \mathrm{d}\left(3 x^2-1\right)=\dfrac{1}{6} \int_{-1}^{11} f(t) \mathrm{d} t$ (với) $t=3 x^2-1$.
$\Leftrightarrow I_2=3$.
Vậy $I=I_1+I_2=7$.
+) $I_1=\int_0^2 2 x \mathrm{~d} x=\left.x^2\right|_0 ^2=4$.
+) $I_2=\int_0^2 x . f\left(3 x^2-1\right) \mathrm{d} x=\dfrac{1}{6} \int_0^2 f\left(3 x^2-1\right) \mathrm{d}\left(3 x^2-1\right)=\dfrac{1}{6} \int_{-1}^{11} f(t) \mathrm{d} t$ (với) $t=3 x^2-1$.
$\Leftrightarrow I_2=3$.
Vậy $I=I_1+I_2=7$.
Đáp án D.