Question

How do you evaluate the integral `int dx/sqrt(3-x)` from 1 to 3 if it converges?

Answer 1

`2sqrt2`

Explanation:

We have:

`int_1^3dx/sqrt(3-x)`

We should try substitution, namely by letting `u=3-x`. This implies that `du=-dx`. The bounds will also change, with `1` becoming `3-1=2` and `3` becoming `3-3=0`.

`int_1^3dx/sqrt(3-x)=-int_2^0(du)/sqrtu`

We can use fractional and negative exponents in the integrand, as well as switching the direction of the bounds since we have the negative sign out front.

`-int_2^0(du)/sqrtu=int_0^2u^(-1/2)du`

Integrate using the rule: `int_a^bu^ndu=[u^(n+1)/(n+1)]_a^b`

`int_0^2u^(-1/2)du=[u^(-1/2+1)/(-1/2+1)]_0^2=[u^(1/2)/(1/2)]_0^2=[2sqrtu]_0^2=2sqrt2`

Answer 2

`2 sqrt 2`

Explanation:

`int_1^3 1/sqrt(3-x) \ dx`

we have a pattern here from the power and chain rules

`d/dx(sqrt(3-x) )`
`= 1/2 * 1/sqrt(3-x) * (-1) = -(1/2)/sqrt(3-x)`

so our integrand is `- 2 d/dx(sqrt(3-x) )`

and so we are integrating

`int_1^3 \ - 2d/dx(sqrt(3-x) ) dx`

`= - 2 [sqrt(3-x) ]_1^3 = 2 [sqrt(3-x) ]_3^1`

`= 2( sqrt 2 - 0) = 2 sqrt 2`