Home > Calculus > Integration: the Area Problem

How do you find the integral of #int dx/(sqrt(2x-1)# from 1/2 to 2?

Answer 1

Emilia Aguilar

Try the substitution #u = 2x-1# in hopes of getting #int u^(-1/2) du#.

With #u = 2x-1#, we get #du=2dx# so the indefinite integral becomes

#int (2x-1)^(-1/2) dx = 1/2 int u^(-1/2) du#

The limits of integration change from #x=1/2# to #u=2(1/2)-1 = 0# and from #x=2# to #u=2(2)-1=3#

The new problem is to evaluate

#1/2int_0^3 u^(-1/2) du#

# = [u^(1/2)]_0^3 = sqrt3#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Cameron Alexander

To find the integral of ( \int_{1/2}^{2} \frac{dx}{\sqrt{2x-1}} ), we can use the substitution method. Let ( u = 2x - 1 ), then ( du = 2dx ).

After substitution, the integral becomes ( \int_{1}^{3} \frac{du}{2\sqrt{u}} ).

This simplifies to ( \frac{1}{2} \int_{1}^{3} u^{-1/2} ).

Integrating ( u^{-1/2} ) with respect to ( u ) yields ( 2u^{1/2} = 2\sqrt{u} ).

Evaluating this from 1 to 3, we get ( 2\sqrt{3} - 2\sqrt{1} = 2\sqrt{3} - 2 ). Therefore, the integral of ( \int_{1/2}^{2} \frac{dx}{\sqrt{2x-1}} ) is ( 2\sqrt{3} - 2 ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.