Math 2300 – Calculus II – University of Colorado
Fall 2010 – Midterm I – Solutions

(26 points) Consider the integral

(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:

(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:

(4 points) The formula for error in an approximation is given by
Compute the errors in Left(2) and Mid(2).
Solution:

(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:

(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:
