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A final exam example for Calculus II (UN1102) Section 2. The exam consists of eight problems, each worth 20 points. The exam covers topics such as variable substitution, integration by parts, Pythagorean type-formulae for trigonometric functions, basic integrals, quadratic formula, length, area of a surface of revolution, and volumes. The problems in the exam cover topics such as finding the area lying inside two circles, finding the volume of a solid obtained by rotating a region under a curve, finding the value of a constant for which an improper integral converges, and finding the radius of convergence of a series.
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Name:
UNI:
Please write your name and UNI above. The exam consists of eight problems, each worth 20 points. No calculators are allowed in the exam. Please write neatly, and please justify your answers. You are free to use any trigonometric identities that you remember without justification. Partial credit will be given for partial attempts. Full credit will not be given for answers without justification.
Grade table: (teacher use only)
Problem 1 : Problem 2 : Problem 3 : Problem 4 : Problem 5 : Problem 6 : Problem 7 : Problem 8 :
Variable substitution:
x = g(u); dx = g′(u)du;
f (x) dx =
f (g(u))g′(u) du.
Another variant:
u = g(x); du = g′(x)dx;
f (g(x))g′(x) dx =
f (u) du.
Integration by parts: ∫ udv = uv −
vdu.
Pythagorean type-formulae for trigonometric functions:
sin^2 x + cos^2 x = 1; tan^2 x + 1 = sec^2 x.
(The above are useful for removing quantities like
a^2 ± x^2 from integrals via a trig substitution!) Sin angle addition formula:
sin(x + y) = sin(x) cos(y) + sin(y) cos(x)
Some very basic integrals: ∫ xa^ dx =
a + 1
xa+1, unless a = −1 in which case
x−^1 dx = ln |x|.
∫ sin x dx = − cos x,
cos x dx = sin x
Quadratic formula:
Solutions to ax^2 + bx + c = 0 are x =
−b ±
b^2 − 4 ac 2 a Length:
ds =
dx^2 + dy^2 .L =
ds =
dy dx
dx =
dx dy
dy.
Area of a surface of revolution
A =
2 πrds
When the axis of revolution is the x axis and the curve is a graph over the x axis the above is
∫ 2 πy(x)
dy dx
dx
and when the axis of revolution is the x axis and the curve is a graph over the y axis the above is
∫ 2 πy
dx dy
dy.
Volumes (both formulae for surface of revolution about x axis):
Power series:
n=
cnxn
Taylor series:
f (x) '
n=
f (n)(c) n!
(x − c)n
Some examples of Taylor series:
ex^ =
n=
xn n!
cos x =
n=
(−1)n^
x^2 n (2n)!
tan−^1 (x) =
n=
(−1)n^
x^2 n+ 2 n + 1
− ln(1 + x) =
n=
(−1)n^
xn n
(1 + x)k^ = 1 + kx + k(k − 1) 2!
x^2 + k(k − 1)(k − 2) 3!
x^3 +... =
n=
k(k − 1)... (k − n + 1) n!
xn
Problem 1. (20 points) Consider a circle A of radius 1 centered at (0, 0) and another circle B of radius 0 < r < 1 centered at (0, 1). Compute the area lying inside B but outside A. Draw a picture of this configuration of circles and shade the area to be computed.
Problem 3. (20 points) Find the value of the constant C for which the improper integral ∫ (^) ∞
0
x^2 + 4
x + 2 dx
converges. For this value of C, evaluate the integral.
Problem 4. (20 points) Find the length of the curve that is the graph of the function
y(x) =
∫ (^) x
1
t^3 − 1 dt
as x ranges in 1 ≤ x ≤ 4.
Problem 6. (20 points) A student forgot the product rule by mistake, and thought that (f g)′^ = f ′g′. However, in the problem the student was solving, this happened to give the right answer. The function f that the student used was f (x) = ex
2 , and the domain of the problem was the interval (1/ 2 , ∞). Moreover, of course g was not just the zero function – that would have been a silly problem. What is an example of a function g for which the student’s calculation might have given the right answer? Hints: write out a differential equation for g. Solve the differential equation. Partial fractions may help in the final steps.
Problem 7. (20 points) The Fibonacci sequence is defined by the equations
f 1 = 1, f 2 = 1, fn = fn− 1 + fn− 2 for n ≥ 3.
Show that
(a) 1 fn+1fn− 1
fn− 1 fn
fnfn+
(b) ∑∞
n=
fn− 1 fn+