MATH 180 SPRING, 1996

ASSIGNMENT #18

3.3, #2. The units of ∫_{a}^{b} f(t) dt are (meters/second^{2}) × seconds = meters/second.

3.3, #5. The area is given by ∫_{0}^{π} sin x dx. We estimate this integral by taking averages of left and right sums for large n. For example, when n = 200, the left and right sums are 1.992 and 2.008. The area under one arch of sin x is then equal to 2. Note that sin x is not monotonic on [0, π], so it does not follow from general principles that the integral is between the left and right sums for given n. However, sin x is increasing on [0, π/2] and decreasing on [π/2, π], so the integrals ∫_{0}^{π/2} sin x dx and ∫_{π/2}^{π} sin x dx can be estimated to any prescribed degree of accurancy. Note also that the arch is symmetric about the line x = π/2 so that ∫_{0}^{π} sin x dx = 2 ∫_{0}^{π/2} sin x dx.

3.3, #16. a) The average of a function over an interval from a to b is defined to be 1/(b − a) ∫_{a}^{b} f(x) dx. If we graph f(x), we note that it is the unit circle restricted to the first quadrant. The area under the graph is 1/4 of the area of the unit circle, thus, π/4, or approximately 0.79.

b) The area is certainly bigger than 1/2. Draw a line segment connecting (0,1) to (1,0). The area under the line segment is 1/2, and the area under f(x) is greater than 1/2.

3.4, #1. F (b) = F (0) + ∫_{0}^{b} f(x) dx = ∫_{0}^{b} f(x) dx by the Fundamental Theorem since F (0) = 0. The integral ∫_{0}^{b} f(x) dx can be interpreted as the sum of ±-areas of regions bounded by the graph of f(x), the x-axis, and the lines y = i for i = 0, 1, …, b. We find

\begin{matrix}b & 1 & 2 & 3 & 4 & 5 & 6 \\ F (b) & 1 & 1.5 & 1 & 0 & −1 & −1.5 \end{matrix}

3.4, #2. a) ∫_{−3}^{0} f(x) dx = −2.

b) ∫_{−3}^{4} f(x) dx = ∫_{−3}^{0} f(x) dx + ∫_{0}^{3} f(x) dx. + ∫_{3}^{4} f(x) dx = −2 + 2 − A/2 = −A/2.

3.4, #4. F is increasing when F′ is positive, decreasing when F′ is negative. So F is increasing from 0 to about 1.77 and decreasing from about 1.77 to 2.5.