Quiz 3 for MATH 105 SECTION 205
February 04, 2015
Given Name
Family Name
Student Number
2
Z
1. (a) (1 point) Write down the midpoint Riemann sum with n = 50 to approximate
1
notations.
50
X
(a)
k=1
Answer.
Since n = 5, then ∆x =
So the midpoint Riemann sum is: x∗k =
f (x∗k )∆x
=
50
X
k=1
k=1
0.02
1 + 0.02(k − 0.5)
2−1
1
=
= 0.02 and
50
50
for all k = 0, 1, 2, · · · , 5.
xk = 1 + k∆x = 1 + 0.02k,
50
X
1
dx using sigma
x
1 + 0.02(k − 1) + 1 + 0.02k
xk−1 + xk
=
= 1 + 0.02(k − 0.5) and
2
2
50
X
1
0.02
· 0.02 =
.
1 + 0.02(k − 0.5)
1 + 0.02(k − 0.5)
k=1
(b) (0.5 points) Express 2 + 4 + 6 + 8 + · · · + 80 using sigma notations.
40
X
(b)
2k
k=1
Let x1 = 2, x2 = 4, x3 = 6, · · · , xn = 80, it’s easy to see the patten is:
Answer.
xk = 2k,
for all k = 1, 2, · · · , n.
Since xn = 80 = 2n, then n = 40. So we get
2 + 4 + 6 + 8 + · · · + 80 =
40
X
2k.
k=1
Z
(c) (1 point) Let f and g be two integrable functions on [1, 2], if
Z 2
compute
[2f (x) − 3g(x)] dx.
2
Z
f (x) dx = 2 and
1
1
(c)
Answer.
In fact, we have
Z 2
Z
[2f (x) − 3g(x)] dx = 2
1
2
Z
f (x) dx − 3
1
= 2·2−3·1
= 1.
g(x) dx = 1,
1
1
2
g(x) dx
1
2
√
1 − x2 , if 0 ≤ x ≤ 1,
2. Let f (x) =
, and R be the region bounded by the graph of f (x) and x-axis between
2 − 2x,
if x > 1,
x = 0 and x = 3, then
(a) (1 point) Find the area of R.
π
+4
4
(a)
Answer.
The region bounded by the graph of f (x) and x-axis between x = 0 and x = 3 consists of a
1
-disc and a triangle.
4
1
π
1
π
1
-disc, its area is π · 12 = . For the triangle, its area is · 2 · 4 = 4. So the area of R is + 4.
4
4
4
2
4
(b) (1 point) Find the net area of R.
For the
π
−4
4
(b)
1
-disc is above the x-axis, it has positive net area, but the triangle is below the x-axis,
4
π
it has negative net area. So the net are of R is − 4.
4
Z 3
(c) (0.5 points) Compute
f (x) dx.
Answer.
The
0
π
−4
4
(c)
Z
3
Z
Answer.
By the geometric meaning of
f (x) dx, it’s just the net area of R, that is,
0
π
− 4.
4
1, if x is a rational number
3. Let f (x) =
, then
0, if x is not a rational number
(a) (1. points) What’s the value for the left Riemann sum for any regular partition of [0, 1]?.
1
(a)
Answer.
For any regular partition with n, we know that ∆x =
xk = k∆x =
k
,
n
1
and
n
for all k = 0, 1, 2, · · · , n.
3
f (x) dx =
0
It’s easy to see that left end point x∗k =
So the left Riemann sum is:
n
X
k−1
is always a rational number, which implies that f (x∗k ) = 1.
n
f (x∗k )∆x =
k=1
n
X
∆x =
k=1
n
X
1
1
= n · = 1.
n
n
k=1
(b) (0.5 points) For any partition of [0, 1], is it true that we can always find a Riemann sum whose value is
0? (Just put ‘Yes’ or ‘No’).
(b)
Yes
Answer.
For any partition a = x0 < x1 < x2 < · · · < xn = n, for any subinterval [xk−1 , xk ], we can
take an irrational number x∗k which belongs to [xk−1 , xk ], which implies that f (x∗k ) = 0. So we get
n
X
f (x∗k )∆xk =
k=1
n
X
0 · ∆xk = 0.
k=1
(c) (0.5 points) Is f integrable on [0, 1]? (Just put ‘Yes’ or ‘No’).
(c)
No
Answer.
If f is integrable on [0, 1], for any regular partition, when n → 0, we should get the same
limit. But by the results of part a and part b, we could get two limits 1 and 0, contradiction. Therefore,
f is not integrable on [0, 1].
Z 1
4. (3 points) Use the Riemann sum to compute
(2x + 1) dx.
0
Answer.
Since f (x) = 2x + 1 is continuous on [0, 1], so f is integrable on [0, 1]. Hence we can use the
Z 1
right Riemann sum to approximate
(2x + 1) dx. Now for any n, the regular partition using n tells us that
0
1−0
1
∆x =
= and
n
n
xk = 0 + k∆x =
So the right Riemann sum is: x∗k = xk =
n
X
k
n
k
,
n
for all k = 0, 1, 2, · · · , n.
and
f (x∗k )∆x
=
k=1
=
n
X
1
k
·
f
n
n
k=1
n
1 X 2k
·
+1
n
n
k=1
!
n
n
X
2X
=
k+
1
n
k=1
k=1
1
2 n(n + 1)
=
·
·
+n
n
n
2
1
=
[n + 1 + n]
n
2n + 1
=
n
→ 2, as n → ∞.
1
·
n
So we get
Z
1
(2x + 1) dx = 2.
0
Your Score:
/10