Cho hàm số $f\left( x \right)$ có $f'\left( x \right) = \left( {x + 4} \right)\sqrt {x + 1} $ với mọi $x > - 1$ và $f\le?
Cho hàm số \(f\left( x \right)\) có \(f'\left( x \right) = \left( {x + 4} \right)\sqrt {x + 1} \) với mọi \(x > - 1\) và \(f\left( 0 \right) = 2\). Tích phân \(\int_0^3 {f\left( x \right){\rm{d}}x} \) bằng
A. \(\dfrac{{1234}}{{35}}\).
B. \(\dfrac{{1334}}{{35}}\).
C. \(\dfrac{{267}}{7}\).
D. \(\dfrac{{162}}{5}\).
Đáp án B
Chọn BXét \(I = \int_{}^{} {\left( {x + 4} \right)\sqrt {x + 1} {\rm{d}}x} \), đặt \(t = \sqrt {x + 1} \Rightarrow {t^2} = x + 1 \Rightarrow 2tdt = dx\). Khi đó:
\(I = \int_{}^{} {\left( {{t^2} + 3} \right){\rm{t2td}}x} = \int_{}^{} {\left( {2{t^4} + 6{t^2}} \right){\rm{d}}x} = \dfrac{{2{t^5}}}{5} + 2{t^3} + C = \dfrac{{2{{\left( {x + 1} \right)}^{\dfrac{5}{2}}}}}{5} + 2{\left( {x + 1} \right)^{\dfrac{3}{2}}} + C\).
Suy ra \(f\left( x \right) = \dfrac{{2{{\left( {x + 1} \right)}^{\dfrac{5}{2}}}}}{5} + 2{\left( {x + 1} \right)^{\dfrac{3}{2}}} + C\). Thay \(x = 0\): \(f\left( 0 \right) = \dfrac{2}{5} + 2 + C \Rightarrow C = - \dfrac{2}{5}\). Do đó \(f\left( x \right) = \dfrac{2}{5}{\left( {x + 1} \right)^{\dfrac{5}{2}}} + 2{\left( {x + 1} \right)^{\dfrac{3}{2}}} - \dfrac{2}{5}\).
Khi đó \(\int_0^3 {\left( {\dfrac{2}{5}{{\left( {x + 1} \right)}^{\dfrac{5}{2}}} + 2{{\left( {x + 1} \right)}^{\dfrac{3}{2}}} - \dfrac{2}{5}} \right){\rm{d}}x} = \left. {\left( {\dfrac{4}{{35}}{{\left( {x + 1} \right)}^{\dfrac{7}{2}}} + \dfrac{4}{5}{{\left( {x + 1} \right)}^{\dfrac{5}{2}}} - \dfrac{2}{5}x} \right)} \right|_0^3 = \dfrac{{1334}}{{35}}\).
