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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 0 \right)=\dfrac{2}{3}$ và $\left( \sqrt{x}+\sqrt{x+1} \right).{f}'\left( x \right)=1,\forall x\ge -1$. Biết $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{a\sqrt{2}+b}{15}$ với $a,b\in \mathbb{Z}$. Tính $T=a+b$
A. $-8$.
B. $-24$.
C. $24$.
D. $8$.
A. $-8$.
B. $-24$.
C. $24$.
D. $8$.
Ta có $\left( \sqrt{x}+\sqrt{x+1} \right).{f}'\left( x \right)=1$ $\Leftrightarrow {f}'\left( x \right)=\dfrac{1}{\sqrt{x}+\sqrt{x+1}}$ $\Leftrightarrow {f}'\left( x \right)=\sqrt{x+1}-\sqrt{x}$
$\Rightarrow f\left( x \right)=\int{\left( \sqrt{x+1}-\sqrt{x} \right)dx}=\dfrac{2{{\left( x+1 \right)}^{3/2}}}{3}-\dfrac{2{{x}^{3/2}}}{3}+C$
$f\left( 0 \right)=\dfrac{2}{3}\Rightarrow C=0$. $\Rightarrow f\left( x \right)=\dfrac{2{{\left( x+1 \right)}^{3/2}}}{3}-\dfrac{2{{x}^{3/2}}}{3}$.
$\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\dfrac{2}{3}\left[ {{\left( x+1 \right)}^{3/2}}-{{x}^{3/2}} \right]dx}$ $=\left. \dfrac{2}{3}\left[ \dfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\dfrac{2}{5}{{x}^{5/2}} \right] \right|_{0}^{1}=\dfrac{16\sqrt{2}-8}{15}$
$\Rightarrow a=16, b=-8$. Vậy $T=a+b=8$
$\Rightarrow f\left( x \right)=\int{\left( \sqrt{x+1}-\sqrt{x} \right)dx}=\dfrac{2{{\left( x+1 \right)}^{3/2}}}{3}-\dfrac{2{{x}^{3/2}}}{3}+C$
$f\left( 0 \right)=\dfrac{2}{3}\Rightarrow C=0$. $\Rightarrow f\left( x \right)=\dfrac{2{{\left( x+1 \right)}^{3/2}}}{3}-\dfrac{2{{x}^{3/2}}}{3}$.
$\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\dfrac{2}{3}\left[ {{\left( x+1 \right)}^{3/2}}-{{x}^{3/2}} \right]dx}$ $=\left. \dfrac{2}{3}\left[ \dfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\dfrac{2}{5}{{x}^{5/2}} \right] \right|_{0}^{1}=\dfrac{16\sqrt{2}-8}{15}$
$\Rightarrow a=16, b=-8$. Vậy $T=a+b=8$
Đáp án D.