How do you evaluate the integral int dx/sqrt(3-x) from 1 to 3 if it converges?
2 Answers
Explanation:
We have:
int_1^3dx/sqrt(3-x)
We should try substitution, namely by letting
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_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
Explanation:
we have a pattern here from the power and chain rules
so our integrand is
and so we are integrating