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

Question

Ezekiel Alexander · Accepted Answer

#2sqrt2#

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#

Isaiah Allen · Answer

# 2 sqrt 2#

#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#

Emily Ammerman · Answer

undefined To evaluate the integral $ \int_{1}^{3} \frac{dx}{\sqrt{3-x}} $, we first need to check if it converges by examining the integrand at the endpoints of the interval [1, 3]. At $ x = 3 $, the denominator becomes 0, which indicates a potential issue with convergence. However, since the integrand is continuous and does not have a singularity in the interval [1, 3], the integral converges.

To evaluate the integral, we can use a substitution method. Let $ u = 3 - x $, then $ du = -dx $. The limits of integration change as follows: when $ x = 1 $, $ u = 3 - 1 = 2 $, and when $ x = 3 $, $ u = 3 - 3 = 0 $.

The integral becomes $ -\int_{2}^{0} \frac{du}{\sqrt{u}} $, which simplifies to $ -\int_{0}^{2} u^{-1/2} du $. Integrating $ u^{-1/2} $ with respect to $ u $ gives $ -2u^{1/2} $. Evaluating this from 0 to 2, we get $ -2(2^{1/2} - 0) $, which simplifies to $ -2\sqrt{2} $.

Therefore, the value of the integral is $ -2\sqrt{2} $.

Charles Arocha · Answer

undefined To evaluate the integral $ \int_{1}^{3} \frac{dx}{\sqrt{3-x}} $, we can use the substitution method. Let $ u = 3 - x $, then $ du = -dx $. Rewrite the integral in terms of $ u $, then integrate with respect to $ u $. Finally, revert back to the variable $ x $ to find the result.