Câu 14
Ta có
$\int \dfrac{dx}{\sqrt{x+2} + \sqrt{x+1}} = \int \dfrac{\sqrt{x+2} - \sqrt{x+1}}{x + 2 - (x + 1)}dx$
$= \int (\sqrt{x+2} - \sqrt{x+1}) dx$
$= \dfrac{2}{3} \sqrt{(x+2)^3} - \dfrac{2}{3} \sqrt{(x+1)^3} + c$
$= \dfrac{2}{3} (x+2)\sqrt{x+2} - \dfrac{2}{3} (x+1)\sqrt{x+1} + c$
Vậy $a = \dfrac{2}{3}, b = - \dfrac{2}{3}$. Vậy
$S = 3a + b = 2 - \dfrac{2}{3} = \dfrac{4}{3}$
Câu 15
Ta có
$f(x) = \int \dfrac{dx}{\sqrt{x} + \sqrt{x+1}}$
$= \int \dfrac{\sqrt{x+1} - \sqrt{x}}{x + 1 - x} dx$
$= \int (\sqrt{x+1} - \sqrt{x} )dx$
$= \dfrac{2}{3} (x+1)\sqrt{x+1} - \dfrac{2}{3}x\sqrt{x} + c$
Khi đó
$f(0) = c = \dfrac{2}{3}$
Do đó
$f(x) = \dfrac{2}{3} (x+1)\sqrt{x+1} - \dfrac{2}{3}x\sqrt{x} +\dfrac{2}{3}$
Vậy
$f(3) = \dfrac{16}{3} - 2\sqrt{3} + \dfrac{2}{3} = 6-2\sqrt{3}$
và
$f(2) = 2\sqrt{3} - \dfrac{4}{3} \sqrt{2} + \dfrac{2}{3}$
Do đó
$T = 3[ 6-2\sqrt{3} + 2\sqrt{3} - \dfrac{4}{3} \sqrt{2} + \dfrac{2}{3}] + 4\sqrt{2}$
$=18 -4\sqrt{2} + 2 + 4\sqrt{2} = 20$
Vậy $T = 20$.
