7.9, #21. If x ≥ 3, then x^{2} ≥ 3x and e^{−x^{2}} ≤ e^{−3x}. Thus

∫_{3}^{∞} e^{−x^{2}} dx ≤ ∫_{3}^{∞} e^{−3x} dx = −\frac{1}{3} lim_{b→∞} e^{−3x}|_{3}^{b} = \frac{e^{−9}}{3} .

Similarly if x ≥ n, then x^{2} ≥ nx and e^{−x^{2}} ≤ e^{−nx}. Thus

∫_{n}^{∞} e^{−x^{2}} dx ≤ ∫_{n}^{∞} e^{−nx} dx = −\frac{1}{n} lim_{b→∞} e^{−nx}|_{n}^{b} = \frac{e^{−n^{2}}}{n} .

7.9, #22. \frac{2x^{2} + 1}{4x^{4} + 4x^{2} − 2} behaves like \frac{2x^{2}}{4x^{4}} = \frac{1}{2x^{2}} for large x. By the p-test integral (a)

should converge while integral (b) should diverge (since (1/2x^{2})^{1/4} = 1/2^{1/4}x^{1/2}). Indeed, 4x^{2} − 2 ≥ 0 for x ≥ 1, so

∫_{1}^{∞} \frac{2x^{2} + 1}{4x^{4} + 4x^{2} − 2} dx ≤ ∫_{1}^{∞} \frac{2x^{2} + 1}{4x^{4}} dx

= ∫_{1}^{∞} \frac{dx}{2x^{2}} + ∫_{1}^{∞} \frac{dx}{4x^{4}}

= \frac{1}{2} + \frac{1}{12} = \frac{7}{12},

using the antiderivatves −1/2x and −1/12x^{3} to calculate the two integrals. On the other hand,

∫_{1}^{∞}(\frac{2x^{2} + 1}{4x^{4} + 4x^{2} − 2})^{1/4} dx ≥ ∫_{1}^{∞}(\frac{2x^{2}}{4x^{4} + 4x^{4}} )^{1/4} dx

since 2x^{2} + 1 > 2x^{2} and 4x^{4} + 4x^{2} − 2 < 4x^{4} + 4x^{4} for x ≥ 1. The last integral diverges by the p-test.

7.9, #25. The tangent line approximation 1 + t to e^{t} at t = 0 lies below the graph of e^{t} since the graph of e^{t} is concave-up. Thus 1 + t ≤ e^{t} for all t. Substituting t = 1/x in this inequality gives 1 + (1/x) ≤ e^{1/x} or e^{1/x} − 1 ≥ (1/x). Thus

∫_{1}^{∞} \frac{dx}{x^{5}(e^{(1/x)} − 1)} < ∫_{1}^{∞} \frac{x}{x^{5}} dx

The second integral converges by the p-test, so the first one converges as well.

7.10, #2: F (x) starts at (0, 0) and is increasing, concave-up on its initial segment, concavedown on its second segment, where it approaches a horizontal asymptote.

7.10, #3. F (x) starts at (0, 0) is increasing, concave-down on its initial segment, concaveup on its middle segment, concave-down on its last segment.

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7.10, #6. Note the following features of the six slope fields: (I) and (III) have horizontal slopes for large |x|. (II) and (IV) have steep slopes for large |x| with (IV) having almost vertical slopes. Lastly, (V) and (VI) are the only slope fields with negative slopes. It follows easily that (d) e^{−0.5x} cos x is matched with (V) and (f) −e^{−x^{2}} with (VI). The difference between (I) and (III) is that of scale with respect to the x-direction. So (b) e^{−2x^{2}} and (c) e^{−x^{2}/2} are matched respectively with (I) and (III). Finally, (a) e^{x^{2}} is matched with (IV) and (e) (1/(1 + 0.5 cos x)^{2} with (II). Indeed, since 1/2 ≤ 1 + 0.5 cos x ≤ 3/2, it follows that 4/9 ≤ 1/(1 + 0.5 cos x)^{2} ≤ 4. Thus the steepness of the slope field of (e) is bounded, and the match must be (e) with (II) and (a) with (IV).

7.10, #8. \frac{d}{dx} ∫_{0}^{x} \sqrt{3 + cos(t^{2})} dt = \sqrt{3 + cos(x^{2})}.

7.10, #10. \frac{d}{dx} ∫_{0.5}^{x} arctan(t^{2}) dt = arctan(x^{2}).