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

2 Answers
Jul 25, 2016

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

Jul 25, 2016

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