UI Putnam Training Sessions, Beginner Level
Problem Set 11: Integrals, I Solutions
http://www.math.illinois.edu/contests.html
The Problems
1. Warmup integrals. Each of the following integrals has a simple evaluation that requires almost
no calculations if approached the right way. Try to find this simple approach using some of the
tricks mentioned above: symmetry, change of variables, properties of trig functions, etc.
Z
(a)
1
x(1 − x)2014 dx. (Hint: Integrate backwards!)
0
Solution: 1/(2015 · 2016) (Direct computation doesn’t work since the exponent 2014 is on
the “wrong” term, 1 − x, instead of x. Changing variables y = 1 − x effectively switches the
exponent to give an integral over y 2014 (1 − y), which can be easily evaluated by splitting it up
into two.)
Z 2π
(b)
(x − π)2013 (1 + sin2014 x) dx (Hint: Change variables to get an integral over the symmetric
0
interval [−π, π] and exploit symmetry.)
Solution: 0 (Change variables y = x − π. The integrand becomes an odd function of y
(i.e., satisfies f (−y) = −f (y)), and the integration interval turns into the symmetric interval
[−π, π]. Hence the integral is 0.)
2. Contest integrals, I. The Putnam and Virginia Tech Math Contests often include integration
problems that can be done using tricks like those above (symmetries, backwards integration, trig
identities, etc). If you see an integral on a contest, don’t try standard methods (e.g.,
tricky substitutions, integration by parts); these almost never work. Instead, try to
find and exploit symmetries or other special features of the integral.
Here is an example illustrating this (Problem B1 from the 1987 Putnam):
p
Z 4
ln(9 − x)
p
p
dx
ln(9 − x) + ln(x + 3)
2
Solution: 1 We exploit the symmetry of the integrand about x = 3. Let f (x) denote the
integrand.R It is easily checked
R 1 that f (3 − y) + f (3 + y) = 1 for 0 < y < 3. The given integral then
1
becomes −1 f (3 + y)dy = 0 (f (3 − y) + f (3 + y))dy = 1.
3. Contest integrals, II. This one is a bit harder, but it becomes easy once you spot the underlying
symmetry. (Problem A3 from the 1980 Putnam)
Z
0
π/2
dx
1 + (tan x)
1
√
2
UI Putnam Training
Problem Set 11: Integrals, I Solutions
Fall 2016
Solution: π/4 Let I denote the given integral. Writing tan x = sin x/ cos x and clearing denominators gives
√
Z π/2
(cos x) 2
√
√ dx.
I=
(cos x) 2 + (sin x) 2
0
Now, make the change of variables x = π/2 − y. Since cos(π/2 − y) = sin y and sin(π/2 − y) = cos y,
the integral becomes
√
Z π/2
(sin y) 2
√
√ dy.
I=
(sin y) 2 + (cos y) 2
0
R1
Adding the two formulas for I, we get 2I = 0 1dx = π/2, so I = π/4.
4. Contest integrals, III. This one looks particularly intimidating, but the same sort of tricks work.
(2012 Virginia Tech Math Contest)
Z
0
π/2
cos4 x + sin x cos3 x + sin2 x cos2 x + sin3 x cos x
dx
sin4 x + cos4 x + 2 sin x cos3 x + 2 sin2 x cos2 x + 2 sin3 x cos x
Solution: π/4 (Use the same idea as in the previous problem: Set x = π/2 − y to swap cos and
sin in the numererator, without changing the value of the integral. Adding the two integrals gives
R π/2
1dx = π/2, so the original integral is half this value, i.e., π/4.
0
Fun Problem of the Week
The Lunch Encounter Problem. Suppose you and your friend decide to meet for
lunch at a restaurant. Whoever gets there first waits for the other person. Assuming
each arrives at a random time point between 12:00 and 12:10, what is the probability
the first person has to wait at most 2 minutes? (Hint: Think in 2 dimensions!)
Solution: 0.36 Let x and y denote the arrival times of the two people, measured in minutes after
12:00, so that x and y are random numbers in the interval [0, 10]. Interpret (x, y) as a point in the
square [0, 10] × [0, 10]. The probability we need to compute can then be interpreted as the probability
that a randomly chosen point (x, y) in this square satisfies (∗) |x − y| ≤ 2. This, in turn, is equal
to A(R)/A(S), where A(...) denotes the area, S is the above square, and R is the part of this square
satisfying the condition (∗). By elementary geometry, we get A(S) = 102 , A(R) = 102 − 2(1/2)82 = 36,
A(R)/A(S) = 0.36.
Happy Problemsolving!
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