Math 181H: Hour Exam 1 (Solutions)

Prof. S. Smith: February 18, 1992

Problem 1: If a radioactive element has a half-life of 810 years, how much of an original 10g is left after 300 years?

… = (10g)2^{− \frac{300}{810}} ≅ 7.736g

Problem 2: Find the derivative of f(x) = tan^{−1}(e^{x}).

… = \frac{1}{1+(e^{x})^{2}} e^{x} = \frac{e^{x}}{1+e^{2x}}

Problem 3: Find

lim_{x→0} \frac{cosh x − 1}{x^{2}}

By 0/0 l’H, = lim_{x→0} \frac{sinh x}{2x}; again 0/0 l’H, = lim_{x→0} \frac{cosh x}{2} = \frac{1}{2} Problem 4: Find ∫ x csc^{2} x dx.

By parts: u = x, dv = csc^{2} x dx, get du = dx, v = − cot x; so = −x cot x + ∫ cot xdx = −x cot x + ln | sin x| + C.

Problem 5: Evaluate ∫_{4}^{∞} xe^{−x^{2}}dx.

… = −\frac{1}{2}e^{−x^{2}}|_{4}^{∞} = −\frac{1}{2}(0 − e^{−16}) = \frac{1}{2e^{16}} .

Problem 6: Find ∫ tan^{3} x sec^{3} x dx.

Using u = sec x, du = sec x tan x dx, tan^{2} x = sec^{2} x − 1, get = ∫ (u^{2} − 1)u^{2} du = ∫ u^{4} du − ∫ u^{2} du = \frac{u^{5}}{5} − \frac{u^{3}}{3} = \frac{sec^{5} x}{5} − \frac{sec^{3} x}{3} + C.

Problem 7: Find

∫ \frac{dx}{\sqrt{x^{2} + 2x + 5}}

Notice completing square, x^{2} + 2x + 5 = (x + 1)^{2} + 2^{2}. So use x + 1 = 2 tan u, dx = 2 sec^{2} u du, \sqrt{x^{2} + 2x + 5} = 2 sec u. So = ∫ \frac{2 sec^{2} u du}{2 sec u} = ∫ sec u du = ln | sec u + tan u| = ln |\frac{\sqrt{x^{2}+2x+5}}{2} + \frac{x+1}{2}| + C

= ln |\sqrt{x^{2} + 2x + 5} + x + 1| + C′ (where C′ is C − ln 2). Problem 8: Find

∫ \frac{x − 11}{x^{2} + 3x − 4}dx

Note x^{2} + 3x − 4 = (x + 4)(x − 1), so partial fractions \frac{A}{x+4} + \frac{B}{x−1} = \frac{x−11}{x^{2}+3x−4} gives A(x − 1) + B(x + 4) = x − 11, and then A + B = 1, −A + 4B = −11. Then add to get 5B = −10, so that B = −2, A = 3. Now = ∫ \frac{3}{x+4} − \frac{2}{x−1} dx = 3 ln |x + 4| − 2 ln |x − 1| = ln \frac{|x+4|^{3}}{|x−1|^{2}} + C.

Problem 9: Approximate ∫_{1}^{3} x^{4} dx by Simpson’s rule with n = 4 divisions. (OK to leave answer in fractions).

Have Δx = \frac{3−1}{4} = \frac{1}{2}. (Notice exact value is \frac{x^{5}}{5} |_{1}^{3} = \frac{1}{5}(243 − 1) = 48\frac{2}{5}). So \frac{1/2}{3}[1^{4} + 4(\frac{3}{2})^{4} + 2 · 2^{4} + 4(\frac{5}{2})^{4} + 3^{4}] = \frac{1}{6}[1 + \frac{81}{4} + 32 + \frac{625}{4} + 81] = \frac{1}{6} \frac{581}{2} = \frac{581}{12} = 48\frac{5}{12}.