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Câu hỏi: Cho $\int\limits_{1}^{2}{f\left( x \right)}dx=2$.Khi đó $\int\limits_{1}^{4}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}}$
A. 1.
B. 4.
C. $P=2a+b+c$
D. $P=2a-b+c~$
A. 1.
B. 4.
C. $P=2a+b+c$
D. $P=2a-b+c~$
Lời giải
Đặt x $\sqrt{x}=t\Rightarrow \dfrac{1}{2\sqrt{x}}dx=dt\Rightarrow \dfrac{1}{\sqrt{x}}dx=2dt$
Khi $x=1$ thì $t=1;x=4$ thì $t=2.$
Suy ra $\int\limits_{1}^{4}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}}dx=\int\limits_{1}^{2}{f\left( t \right)}.2dt=2\int\limits_{1}^{2}{f\left( t \right)dt=2.2=4}$
Vậy $\int\limits_{1}^{4}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}}dx=4$
Đặt x $\sqrt{x}=t\Rightarrow \dfrac{1}{2\sqrt{x}}dx=dt\Rightarrow \dfrac{1}{\sqrt{x}}dx=2dt$
Khi $x=1$ thì $t=1;x=4$ thì $t=2.$
Suy ra $\int\limits_{1}^{4}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}}dx=\int\limits_{1}^{2}{f\left( t \right)}.2dt=2\int\limits_{1}^{2}{f\left( t \right)dt=2.2=4}$
Vậy $\int\limits_{1}^{4}{\dfrac{f\left( \sqrt{x} \right)}{\sqrt{x}}}dx=4$
Đáp án B.