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