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Câu hỏi: Tính tích phân $I=\int_0^2 \sqrt{4 x+1} \mathrm{~d} x$.
A. $\dfrac{13}{3}$.
B. 4 .
C. $\dfrac{4}{3}$.
D. 13 .
A. $\dfrac{13}{3}$.
B. 4 .
C. $\dfrac{4}{3}$.
D. 13 .
$
\text { Ta có } I=\int_0^2 \sqrt{4 x+1} \mathrm{~d} x=\int_0^2(4 x+1)^{\dfrac{1}{2}} \mathrm{~d} x=\left.\dfrac{1}{4} \cdot \dfrac{2}{3}(4 x+1)^{\dfrac{3}{2}}\right|_0 ^2=\dfrac{13}{3} \text {. }
$
\text { Ta có } I=\int_0^2 \sqrt{4 x+1} \mathrm{~d} x=\int_0^2(4 x+1)^{\dfrac{1}{2}} \mathrm{~d} x=\left.\dfrac{1}{4} \cdot \dfrac{2}{3}(4 x+1)^{\dfrac{3}{2}}\right|_0 ^2=\dfrac{13}{3} \text {. }
$
Đáp án A.