Section 6.1

Please post comments if you find any errors, or if you wish to contribute answers for additional problems. You may find information on using \LaTeX here.

1. (a) f'(x) = 6e^{2x}

(c) h'(z) = 3z(15z^3 - 30z + 2)e^{5z^3}

(e) \displaystyle{g'(z) = \frac{3}{2}(1 - x)e^{-x}}

(g) \displaystyle{f'(s) = \frac{(1 + 6s)e^{-2s} + 6}{(e^{-2s} + 2)^2}}

2. (a) \displaystyle{\int 3e^{2x}dx = \frac{3}{2}e^{2x} + c}

(c) \displaystyle{\int 4te^{3t}dt = \frac{4}{3}te^{3t} - \frac{4}{9}e^{3t} + c}

(e) \displaystyle{\int z^2 e^z dz = z^2e^z - 2ze^z + 2e^z + c}

(g) \displaystyle{\int e^x\cos(x)dx = \frac{1}{2}(\cos(x) + \sin(x))e^x + c}

3. The maximum value of f is 4e^{-2} at x = 2 .

5. (a) \displaystyle{\lim_{x \to \infty}xe^{-x} = 0}

(c) \displaystyle{\lim_{t \to 0}\frac{e^{-t} - 1}{t} = -1}

8. (a) \displaystyle{\int_0^\infty e^{-x}dx = 1}

(c) \displaystyle{\int_0^\infty xe^{-x}dx = 1}

(e) \displaystyle{\int_0^\infty x^2e^{-x}dx = 2}

10. (a) For -\infty < x < \infty ,

\displaystyle{e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots .}

(cb For -\infty < x < \infty ,

\displaystyle{\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)n!} = \frac{2x}{\sqrt{\pi}} - \frac{2x^3}{3\sqrt{\pi}} + \frac{x^5}{5\sqrt{\pi}} - \frac{x^7}{21\sqrt{\pi})} + \frac{x^9}{108\sqrt{\pi}} - \cdots .}

(c) \mathrm{erf}(1) \approx P_{13}(1) = 0.84271 , rounded to 5 decimal places

11. (a) 26.4 million, rounded to one decimal place

(b) 2072

Leave a Reply