Question

$\int\limits^9_0 dx/1+√x ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,


YOUR QUESTIONS IS--------------$\sf{\int\limits^9_0 dx/(1 + \sqrt{x})}$ Evaluate the given integral expression:


so, put x = t²

dx/dt = 2t

dx = 2t*dt
also, [x = 9 , t = 3] and [x = 0 , t = 0]

therefore,

$\sf{\int\limits^3_0 2t*dx/(1 + t)}$

$\sf{=2\int\limits^3_0 [(t + 1) - 1]/(1 + t). dx}$

$\sf{=2\int\limits^3_0 [dt - dt/(1 + t)]}$

$\sf{=2[t]^3_0 - 2[log|1 + t|]^3_0}$

$\sf{=2[3 - 0] - 2[log4 - log1]}$

$\sf{=6 - 2log2^2}$

$\sf{I = 6 - 4log2}$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int\limits^9_0{\frac{1}{1+\sqrt{x}}}\,dx$

Let √x = P...............(1)
differentiate both sides,
1/2√x dx = dP
dx = 2√x dP
dx = 2P.dP ................(2)
now, upper limit = √9 = 3
lower limit = 0

$\int\limits^3_0{\frac{1}{1+P}}\,2P.dP$

=$2\int\limits^3_0{\frac{P}{1+P}}\,dP$

=$2\int\limits^3_0{\frac{1+P-1}{1+P}}\,dP$

=$2\int\limits^3_0{\frac{1+P}{1+P}-\frac{1}{1+P}}\,dP$

=$2\int\limits^3_0{1-\frac{1}{1+P}}\,dP$

=$2[\int\limits^3_0{dP}-\int\limits^3_0{\frac{1}{1+P}}\,dP$

=$2[P]^3_0-2[log(1+P)]^3_0$

=$6-2[log4]$

= 6 - 4log2