AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
3. The third value in L3 is greater than the second since the integrand is positive
for all X in the interval [0.1, 0.2].
4. 1.8
1
5. . 9 ⎛ 2
T ⎞ 1
cos dT
. 8 ⎛ T
⎜ ⎟ < cos⎜
0 2 0
⎝ ⎠
⎝ 2
2
⎞
⎟ dT
⎠
∫ ∫ because the integrand is negative for all T in the
interval [1.8,1.9]. Thus the function is decreasing.
6. 3.1
In questions 7 to 10, students should observe that the function Y2 has a maximum where
the integrand changes sign from positive to negative and a minimum where the integrand
changes sign from negative to positive. Again, this connection is meant to recall students’
experience with the locations of maximum and minimum points and their relationship
with the sign of the derivative. Ask questions like, “Have you ever seen a situation in
which one ‘thing’ reaches a maximum where another ‘thing’ changes sign from positive to
negative? What are the two ‘things’?”
Students make the connection explicit in question 15.
7. (A) 1.77245 and (B) 3.06998
8. At the first X-intercept, 1.77245, the integrand changes sign from positive to negative.
Thus the function Y2 has a local maximum there. We stop accumulating positive
values of Y1, and start accumulating negative values at that point. Similarly, at
X = 3.06998, the integrand changes sign from negative to positive. Thus we stop
accumulating negative values and start accumulating positive values at that point,
producing a local minimum on the graph of Y2.
9. 3.963327
10. Y2 has a maximum there. Looking at the table of values for L1, L2, and L3, this
occurs near X = 4.
AP® Calculus: 2006–2007 Workshop Materials 71