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

2 Answers
mason m
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#

Eddie
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#

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