Evaluate the integral

Question

Evaluate the integral 

∫(-9/³√(x-9))dx with upper bound 9 and lower bound 1

Answer 1

We can rewrite the problem as

 

-9∫[(x - 9)^-1/3]dx

 

First we will find the integral before plugging in the bounds.  We will do this by the u-substitution method which uses the chain rule.

 

Let u = x - 9

     du = dx

 

 

Next, we substitute this into the original integration.

 

-9∫u^-1/3 du

 

 

Now we can easily integrate.

 

-9[(3/2)u^2/3]+ C

 

 

Substituting u back into this new form, we have

 

-9[(3/2)(x - 9)^2/3]+ C

 

 

Last step is to the plug in the bounds.

 

F(9) - F(1)    where function F is the integral.

 

[-9*((3/2)(9 - 9)^2/3)] - [-9*((3/2)(1 - 9)^2/3)] =

 

[-9*((3/2)(0)^2/3] - [-9*((3/2)(-8)^2/3] =

 

            0           +  9*((3/2) ³√(-8)²) =

 

                             9*((3/2) ³√(64)) =

 

                             9*((3/2)(4)) =

 

                             9(6) =

 

                              54