Answer please?

#int_0^2x/(sqrt(1+x^2))dx#

3 Answers
Nov 26, 2017

#int_0^2 x/sqrt(1+x^2)*dx=sqrt5-1#

Explanation:

#int_0^2 x/sqrt(1+x^2)*dx#

=#int_0^2 (2x*dx)/(2sqrt(1+x^2))#

After using #u=x^2# and #du=2x*dx# transforms, this integral became

#int_0^4 (du)/(2sqrt(1+u))#

=#1/2*int_0^4 (1+u)^(-1/2)*du#

=#1/2*[(1+u)^(-1/2+1)/(-1/2+1)]_0^4#

=#1/2*[(1+u)^(1/2)/(1/2)]_0^4#

=#[(1+u)^(1/2)]_0^4#

=#5^(1/2)-1^(1/2)#

=#sqrt5-1#

Nov 26, 2017

#int_0^2x/(sqrt(1+x^2))dx = sqrt5-1#

Explanation:

let #u = x^2+1#, then #du=2xdx# or #xdx = 1/2du#

Change the limits:

#a = 0^2+1#

#a = 1#

#b = 2^2+1#

#b = 5#

#int_0^2x/(sqrt(1+x^2))dx = 1/2int_1^5u^(-1/2)du#

#int_0^2x/(sqrt(1+x^2))dx = (u^(1/2)]_1^5#

#int_0^2x/(sqrt(1+x^2))dx = sqrt5-1#

Nov 26, 2017

#int_0^2 x/sqrt(1+x^2)dx=sqrt5-1#

Explanation:

We can make a substitution to solve this integral:

Let #u=1+x^2#

#int_0^2 x/sqrt(u)dx#

And so #du=2xdx# which we can then solve for #x# to get #1/2du=xdx#

Replacing #xdx# with #1/2du# we now have

#int_0^2 1/(2sqrt(u))du#

Take out the constant...

#1/2int_0^2 1/(sqrt(u))du#

Applying the power rule for integration:

#1/2int_0^2 (u)^(-1/2)du#

#1/2* ((u)^((-1/2)+1))/((-1/2)+1)#

#1/2* (u)^(1/2)/(1/2)#

#1/2* 2sqrtu#

Simplify and substitute back #u=1+x^2#

#[sqrt(1+x^2)+"C"]_0^2#

Now evaluating the integral from #0# to #2#

#[sqrt(1+color(red)2^2)+"C"] - [sqrt(1+color(red)0^2)+"C"]#

#[sqrt(5)] - [1]#

So

#int_0^2 x/sqrt(1+x^2)dx=sqrt5-1#