Question

Answer please?

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

Answer 1

`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`

Answer 2

`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`

Answer 3

`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`