Midterm 1 Solutions

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Math 2300 – Calculus II – University of Colorado

Fall 2010 – Midterm I – Solutions

1. (26 points) Consider the integral

1. (6 points) Draw a picture of the area that this integral represents. Evaluate this integral any way you like.

Solution: The graph of the function for in represents the upper right quarter (first quadrant) of the circle centered at the origin and of radius . We now want to compute the integral:

Solution 1: Since the integrand is positive on , the integral represents the area under the graph of and above the -axis; this is exactly the area of a quarter of the disc, which is .

Solution 2: We make the trig substitution, and write in as , with in . Then the radical becomes: . The integration limits for are and , and . So:

2. (6 points) Compute Left(2) and Mid(2). Show your work. An answer that appears to have come from a calculator will receive 0 points.

Solution:

3. (4 points) The formula for error in an approximation is given by

Compute the errors in Left(2) and Mid(2).

Solution:

4. (6 points) Estimate the errors in Left(20) and Mid(20) without actually computing these values. Again, an answer that appears to have come from a calculator will receive 0 points.

Solution:

5. (4 points) Order the following approximations from least to greatest for any integer .

Solution:

The function is decreasing, so .

The function is concave down, so .

Since Mid(n) and Trap(n) are generally closer estimates than Left(n) and Right(n), we can order:

2. (12 points)

1. (8 points) Prove that

Solution: Start by rewriting:

Use an integration by parts with and . Then and , so:

Bring to the left side and get:

Adding the terms together on the left:

Dividing both sides by the coefficient , it follows immediately that:

2. (4 points) Use the formula in (a) to evaluate .

Solution: We apply the formula at (a) for :

3. (12 points)

1. (6 points) Show that

Solution: We start by setting up the limit for the improper integral:

We use integration by parts, with and . Then and (using a substitution ) .

But

Since both top and bottom of the fraction go to when , we can use L'Hôpital's rule. Calculate the limit for the ratio of the derivatives:

Hence the original limit is zero, which means that:

2. (6 points) Show that

Solution: Start with the right side: make a substitution that will transform into . That is, . Then , and the new integration limits are and .

4. (12 points)

1. (6 points) Let be a continuous function. Prove that

Solution: Start with the right side, for example, and make the substitution . Then and the new integration limits are and :

2. (6 points) Let be a continuous function and let . Prove that

Solution:

5. (12 points) Assume converges and that is a positive constant.

1. (6 points) Does converge? Explain in 3 sentences or less.

Solution 1: Make the substitution . Then:

So converges, since it differs only by a finite value from the original integral , which is known to be convergent.

Solution 2: For , the transformation from to is a shift to the left (notice that the interval is still contained in the domain of the new function). This horizontal shift does not affect the convergence of the integral, only the finite limit value.

2. (6 points) Does converge as well? Explain in 3 sentences or less.

Solution 1:

But . Since we known that is convergent, their sum will be blow up to , making divergent.

Solution 2: The function is a shift up by units. This will prevent the proper decay of , and thus introduces an additional positive infinite area between the x-axis and the graph of the integrand, preventing it from converging. So diverges.

6. (12 points) Some values of the continuous, differentiable function are given in the table below.

1 5/4 3/2 7/4 2
2 3 4 7 10
1. (6 points) Estimate the integral using these data.

Solution: We can use, for example, the trapezoidal sum for :

2. (6 points) Estimate the integral using these data.

Solution:

7. (14 points) The Law of Mass Action tells us that the time, , taken by a chemical to create a quantity of the product (in molecules) is given by

where and are initial quantities of the two ingredients used to make the product, and is a positive constant. Suppose .

Let's first calculate using partial fractions:

We want to find expressions for and (that will depend on the constants , and ) so that:

Plug in and obtain , i.e. .

Plug in and obtain , i.e. .

Then

1. (8 points) Find the time taken to make a quantity of the product.

Solution:

2. (6 points) What happens to as ?

Solution:

Since , it means that . So: . In other words, is the equilibrium quantity.