Math 105/206 - Quiz 2, Feb 6 2015
IMPORTANT: Write your name AND student number somewhere on this sheet.
No calculators, books, notes. Please show your work to get full marks. (10 marks total +2 “bonus” marks)
If you get more than 10 marks, the excess ones will be transferred to other quizzes.
Problem 1
Given the function f (x) = 3x − 2 :
(a) Compute the right Riemann sum
R 2of f , with n = 4 subintervals and relative to the interval [0, 2] (2 marks).
(b) Compute the definite integral 0 f (x)dx by using its interpretation as a net area (don’t use the fundamental theorem of calculus) (2 marks).
R2
(2 bonus marks) Compute the definite integral 0 f (x)dx by writing it as a limit of right Riemann sums
P
with n subintervals of equal length. You will need the formula nk=1 k = n(n+1)
. Your final result should be a
2
number (the same one you get from point (b) above).
Solution
(a) If we subdivide [0, 2] in 4 subintervals, these will be the four intervals [0, 12 ], [ 12 , 1], [1, 23 ], [ 32 , 2], so the
length of each subinterval is ∆x = 12 , and we are calculating f in the right endpoints. The values of f in the
right endpoints are
3
1
5
1
f (1) = 1,
f
f (2) = 4
=− ,
= ,
f
2
2
2
2
The sum we have to compute is
1 1
1 5 1
1
7
− · +1· + · +4· = .
2 2
2 2 2
2
2
(b) By drawing the graph of f (a line), we see that it intersects the x-axis in x = 23 , which is inside the
interval [0, 2], and that f (x) is negative for x < 23 and positive for x > 23 . The regions that contribute to the
integral are two right triangles. The area of the triangle on the right is A = 12 (2 − 32 )f (2) = 83 , and the area of
the one on the left is B = | 12 23 f (0)| = 23 . So
Z 2
8 2
6
f (x)dx = A − B = − = = 2.
3 3
3
0
(bonus question)
If we subdivide [0, 2] in n subintervals, then the k-th subinterval [xk−1 , xk ] will have length
endpoint’s x-coordinate is 2k
. So we have
n
Z 2
n X
2k
2
3
−2 · .
f (x)dx = lim
n→∞
n
n
0
k=1
Now by the constant multiple and sum rules we get
n X
2k
2
3
−2 · =
n
n
k=1
=
and taking the limit for n → ∞ we get
12
2
n
2 6 X
· ·
k
n n k=1
!
−4
12 n(n + 1)
·
−4
n2
2
− 4 = 6 − 4 = 2.
2
n
and right
Problem 2
(a) Compute the indefinite integrals (3 marks)
Z
Z
3x3 − 4x2 + 1
dx
(cos(x) + 2 sin(3x))dx
x2
Z
e4x dx.
(b) Use the fundamental theorem of calculus and the results of (a) to compute the definite integrals (3
marks)
Z 1
Z 2π
Z 2 3
3x − 4x2 + 1
(cos(x) + 2 sin(3x))dx
e4x dx.
dx
2
x
π
0
1
Solution
(a) We have
Z
3x3 − 4x2 + 1
dx =
x2
3x3 4x2
1
− 2 + 2 dx
x2
x
x
Z
Z
Z
= 3xdx − 4dx + x−2 dx
Z 3
= x2 − 4x − x−1 + C
2
Z
Z
Z
2
(cos(x) + 2 sin(3x))dx = cos(x)dx + 2 sin(3x)dx = sin(x) − cos(3x) + C
3
Z
1
e4x dx = e4x + C.
4
(b) Using the formulas of part (a) we get
Z
1
Z
π
2π
2
2
3 2
1
3
3x3 − 4x2 + 1
12
−1 dx = x − 4x − x =
−8− −
−4−1 =1
x2
2
2
2
2
1
2π 2
2
2
4
(cos(x) + 2 sin(3x))dx = sin(x) − cos(3x) = 0 −
− 0 − (−1) = −
3
3
3
3
π
1
Z 1
1 1
1
e4x dx = e4x = e4 − .
4
4
4
0
0