MATH 116 PRACTICE MIDTERM 2

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.

PROBLEM 1  Integrate the following functions

A)                        

Solution

We use integration by parts here

          u  =  ln(2x)    du  = 1/x dx       from 2(1/2x)
    dv  =  x         v  =  x² / 2

The integration by parts formula gives

B)                   

Solution

For this integral, we use substitution with

        u  =  1 - x        du  =  -dx        x  =  1 - u        1 + x  =  2 - u

This gives

C)                  

 Solution

We use substitution again

        u  =  5x⁵ - 3    du  =  25x⁴ 

This gives

PROBLEM 2  Find the area of the region bounded by the curve

        y  =  x³ - 3x² - 4x  

and the line

        y  =  6x

 Solution

We first sketch the graph of the region

We can see that we will need to do two integrals since on the left (R₁) the top curve is the cubic and the bottom curve is the line and on the right hand side (R₂) the top curve is the line and bottom is the cubic.  From the graph it appears that the equations intersect at -2, 0, and 5.  Let's verify this.  We set the equations equal to each other.

        x³ - 3x² - 4x    =  6x             x³ - 3x² - 10x  =  0

        x(x² - 3x - 10)  =  0            x(x + 2)(x - 5)  =  0  

As we guessed the intersection points are at -2, 0, and 5.  Now we integrate

PROBLEM 3  Approximate the following integral using the midpoint rule with n  =  4.

  Solution

We first calculate

                       b - a             14 - 2        
        Dx  =                    =                  =  3
                         n                    4

The we get the four rectangles with left and right endpoints

        [2,5],   [5,8],   [8,11],   [11,14]

The midpoints of these rectangles are 

        3.5, 6.5, 8.5, and 12.5

Now we use the midpoint rule.  The integral can be approximated by

PROBLEM 4  The revenue R in thousands of dollars at time t years after 1990 of a travel agency can be modeled by the equation  R(t)  =  t² + t e^-t .  Find the total revenue of the company from 1990 to 2000.

Solution

The total revenue is the integral 

We use integration to integrate the second term

        u  =  t        dv  =  e^-t 
        du  =  dt    v  =  -e^-t 

This gives us

PROBLEM 5  Set up the integral to find the volume of the solid that results when the region bounded by the curves

         y  =  3 - x  and y  =  3 + x - x² 

is revolved about the x-axis.  You do not need to evaluate the integral.

We first sketch the graph.  Then draw a cross-section and revolve it about the x-axis.  

We see that the cross-section is a washer with larger and smaller radii 

        R  =  3 + x - x²        and        r  =  3 - x

The area of the cross-section is

        A  =  p(R² - r²)  =  p[(3 + x - x²)² - (3 - x)²]

The curves appear to intersect at 0 and 2.  We solve to make sure of this.

        3 + x - x²  =  3 - x        x² - 2x  =  0

        x(x- 2)  =  0

and they do intersect at 0 and 2.  Putting this together gives