6.2, #4. |x| is non-negative, so ∫_{−1}^{1} |x| dx is the area of two triangles: the first has sides on the lines y = −x, x = −1, y = 0; the second has sides on the lines y = x, x = 1, y = 0. Clearly ∫_{−1}^{1} |x| dx = 1

6.2, #5. a) \frac{1}{\sqrt{2π}} ∫_{1}^{3} e^{− \frac{x^{2}}{2}} dx = \frac{1}{\sqrt{2π}} ∫_{0}^{3} e^{− \frac{x^{2}}{2}} dx−

\frac{1}{\sqrt{2π}} ∫_{0}^{1} e^{− \frac{x^{2}}{2}} dx = 0.4987 − 0.3413 = 0.1574.
b) \frac{1}{\sqrt{2π}} ∫_{−2}^{3} e^{− \frac{x^{2}}{2}} dx = \frac{1}{\sqrt{2π}} ∫_{−2}^{0} e^{− \frac{x^{2}}{2}} dx +\frac{1}{\sqrt{2π}} ∫_{0}^{3} e^{− \frac{x^{2}}{2}} dx
= \frac{1}{\sqrt{2π}} ∫_{0}^{2} e^{− \frac{x^{2}}{2}} dx +\frac{1}{\sqrt{2π}} ∫_{0}^{3} e^{− \frac{x^{2}}{2}} dx = 0.4772 + 0.4987 = 0.9759.

6.3, #2. F′(x) = f(x) is negative on [0, 1] and positive on [1, ∞), so F (x) is decreasing on (0, 1) and increasing on (1, ∞). F′′(x) = f′(x) = 1 is always positive, F (x) is concave up on [0, ∞).

6.3, #3. F′(x) = f(x) is positive on [0, ∞), so F (x) is increasing on [1, ∞). F′′(x) = f′(x) is −1 on [0, 1) and 1 on (1, ∞). So F (x) is concave-down on (0, 1) and concave-up (1, ∞).

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6.3, #6 F′(x) = f(x) > 0 for x > 0 so F (x) is increasing. F′(x) = f(x) is also increasing so the graph of y = F (x) is concave up.

6.3, #11. The cork is oscillating relative to the sea floor. Its position above the sea floor is the signed area bounded by the velocity graph. So at point B the cork is at its crest, at point D the cork is at its trough. At points A and C the cork is halfway between its crest and its trough. The acceleration is zero at points A and C, where the tangent line to the velocity graph is horizontal. The function giving the position of the cork above the sea floor has form − cos t + C for some constant C.