Math 181H: Practice Exam 1 NAME: (was Exam2 in old Math 131, Feb 1990) SS#

Problem 1: Find F′(t) if F (t) = arccos \sqrt{2t − 1}…. − \frac{1}{\sqrt{1−(\sqrt{2t−1})^{2}}} \frac{1}{2\sqrt{2t−1}}2 = \frac{−1}{\sqrt{(2−2t)(2t−1)}}

Problem 2: Find ∫_{0}^{1} x^{2}e^{−x}dx.

By parts with u = x^{2}, dv = e^{−x}dx, get du = 2xdx, v = −e^{−x}, so [−x^{2}e^{−x} + 2 ∫_{0}^{1} xe^{−x}dx]_{0}^{1}. By parts again with u = x, dv = e^{−x}dx, get du = dx, v = −e^{−x}dx,

so = [−x^{2}e^{−x} + 2(−xe^{−x} + ∫ e^{−x}dx)]_{0}^{1} = [−x^{2}e^{−x} − 2xe^{−x} − 2e^{−x}]_{0}^{1}

= (−\frac{1}{e} − 2\frac{1}{e} − 2\frac{1}{e}) − (0 + 0 − 2) = 2 − \frac{5}{e}.

Problem 3: Find ∫0^{\frac{π}{4}} \frac{sin x}{cos^{2}x}dx.

Subst: u = cos x so du = − sin x, to get − ∫0^{\frac{π}{4}} \frac{du}{u^{2}} = u^{−1}|0^{\frac{π}{4}} = \frac{1}{\sqrt{2}/2} − \frac{1}{1} = \sqrt{2} − 1.

Problem 4: Find \frac{dy}{dx} if y = e^{(e^{x})}.

Via logarithmic diff: ln y = e^{x}, so \frac{y′}{y} = e^{x}, hence y′ = ye^{x} = e^{e^{x}}e^{x} = e^{e^{x}+x}.

Problem 5: Find \frac{dy}{dx} if y = \frac{(x+1)^{4}(x+5)^{3}}{(x−3)^{8}} . (Hint: logarithmic differentiation)

ln y = 4 ln(x + 1) + 3 ln(x + 5) − 8 ln(x − 3), so \frac{y′}{y} = \frac{4}{x+1} + \frac{3}{x+5} − \frac{8}{x−3},

hence y′ = [\frac{(x+1)^{4}(x+5)^{3}}{(x−3)^{8}} ][\frac{4}{x+1} + \frac{3}{x+5} − \frac{8}{x−3}]. Problem 6: A bacteria culture starts with 500 and after 3 hours there are 8000. Find an expression for the number P (t) of bacteria after t hours.

Given 8000 = 500e^{3k}, so 16 = e^{3k}, hence ln 16 = 3k, or k = \frac{ln 16}{3}. Thus P (t) = 500e^{\frac{ln 16}{3} t} = 500(16)^{\frac{t}{3}} .

Find the number of bacteria after 4 hours. … 500(16)^{\frac{4}{3}} When will the population reach 20,000?

20000 = 500e^{\frac{ln 16}{3} t}, so 40 = e^{\frac{ln 16}{3} t}; so ln 40 = \frac{ln 16}{3}t, or t = 3\frac{ln 40}{ln 16}. Problem 7: Find ∫ ln \sqrt{x}dx.

By parts, u = ln \sqrt{x}, dv = dx gives du = \frac{1}{\sqrt{x}} \frac{1}{2\sqrt{x}} = \frac{dx}{2x}, v = x. So x ln \sqrt{x} − \frac{1}{2} ∫ dx = x(ln \sqrt{x} − \frac{1}{2}) + C.

Problem 8: Find ∫ \frac{dx}{x^{2}\sqrt{16x^{2}−9}}.

Use trig subst: x = \frac{3}{4} sec u, dx = \frac{3}{4} sec u tan u du, \sqrt{x^{2} − \frac{9}{16}} = \frac{3}{4} tan u.

Then = ∫ \frac{\frac{3}{4} sec u tan u du}{( \frac{9}{16} sec^{2} u)(3 tan u}) = \frac{4}{9} ∫ cos u du = \frac{4}{9} sin u = \frac{4}{9} ^{\frac{\sqrt{x^{2}− \frac{9}{16}}}{x}} = \frac{1}{9} \frac{\sqrt{16x^{2}−9}}{x} + C. Problem 9: Find ∫ \frac{x+1}{x^{2}−x}dx.

Partial fractions: From A(x−1)+Bx = x+1 have A+B = 1, −A = 1 so A = −1, B = 2. Then = − ∫ \frac{dx}{x} + 2 ∫ \frac{dx}{x−1} = − ln x + 2 ln(x − 1) = ln \frac{(x−1)^{2}}{x} + C.

Problem 10: Find ∫ csc^{3}xdx.

By parts: With u = csc x, dv = csc^{2} x have du = − csc x cot xdx, v = − cot x. So integral I = − csc x cot x − ∫ cot^{2} x csc xdx = − csc x cot x − ∫ (csc^{2} x − 1) csc xdx

= − csc x cot x − ∫ csc^{3} xdx + ∫ csc xdx = − csc x cot x − I − ln | csc x + cot x|. Then 2I = − csc x cot x − ln | csc x + cot x|, so I = −\frac{1}{2}(csc x cot x + ln | csc x + cot x|).