MATH 101 V2A
January 7th – Practice problems
Solutions
1. Prove that the function f (x) = x is integrable on [0, 1].
Solution: See Friday’s notes – we did this in class.
2. Prove that the function
(
2 if x 6= 0
f (x) =
1 if x = 0
is integrable on [−a, a], for every a > 0.
Solution: Since this function is equal to the constant function g(x) = 2, you might
guess that the area under the graph of f (x) is the same as the area under the graph
of g(x). Indeed, this is what we will show. That is, we will show that
Z a
f (x)dx = 4a
−a
for every a > 0.
To do this, we first break the interval [−a, a] into n equal subintervals [x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ].
To show that f (x) is integrable on [−a, a] with integral 4a, we need to show that
lim
n→∞
n
X
f (x∗i )∆x = 4a
i=1
for every choice of sample points x∗1 , x∗2 , . . . , x∗n . Now x = 0 will lie in at most two
subintervals (it will lie in only one if it is not an endpoint of a subinterval, but it will
lie in two if it is). This means that at least n − 2 rectangles will have height 2 and the
other 2 rectangles will have a maximum height of 2 and a minimum height of 1. So,
n
4a(n − 1)
2a
2a
2a X
=2·
· (n − 2) + 1 ·
+1·
≤
f (x∗i )∆x ≤ 4a
n
n
n
n
i=1
for any choice of sample points x∗1 , x∗2 , . . . , x∗n . Taking the limit as n goes to infinity of
the above tells us that
n
X
lim
f (x∗i )∆x = 4a
n→∞
for every choice of sample points
integral 4a for any a > 0.
i=1
x∗1 , x∗2 , . . . , x∗n .
2
So, f (x) is integrable on [−a, a] with