6.1, #1. The left and right sums with four subdivisions for ∫_{1}^{2} x^{2} dx are

\begin{aligned}LS & = (1.00)^{2}(.25) + (1.25)^{2}(.25) + (1.50)^{2}(.25) + (1.75)^{2}(.25) = 1.96875 \\ RS & = (1.25)^{2}(.25) + (1.50)^{2}(.25) + (1.75)^{2}(.25) + (2.00)^{2}(.25) = 2.71875 \end{aligned}

Since x^{2} is increasing on [1, 2], it follows that LS ≤ ∫_{1}^{2} x^{2} dx ≤ RS. Moreover, RS − LS = (2^{2} − 1^{2})(.25) = 0.75. In particular, the average (1/2)(LS + RS) = 2.34375 is within (1/2)(0.75) = 0.375 of the actual value.

6.1, #2. Since f(x) is positive on [1, 6], the integral ∫_{1}^{6} f(x) dx is the area bounded between the graph of f, the x-axis, and the lines x = 1, x = 6. So the area is 8.5. The average value of f on [1, 6] is

\frac{1}{6 − 1} ∫_{1}^{6} f(x) dx = \frac{8.5}{5} = 1.7

6.1, #3. The average value of f(x) = 4x + 7 on [1, 3] is

\frac{1}{3 − 1} ∫_{1}^{3}(4x + 7) dx = \frac{1}{2}(2x^{2} + 7x)|_{1}^{3} = 15.

6.1, #5.

∫_{0}^{2}(3x^{2} + 1) dx = (x^{3} + x)|_{0}^{2} = 10.

6.1 #6. Since −1 ≤ sin x ≤ 1 and e^{−x} > 0 for all x, the graph of y = e^{−x} sin x, x ≥ 0, lies between the graphs of y = e^{−x} and y = −e^{−x}. The effect of multiplying sin x by e^{−x} is to ”dampen” the graph of y = sin x. Now the graph of y = sin x on [0, π] is above the x-axis. The portion of the graph on [π, 2π] is the mirror image of that just described and is below the x-axis. Since y = e^{−x} is a decreasing function we now conclude that ∫_{0}^{2π} e^{−x} sin xdx > 0.