Practice Exam
Problem 1 Find
lim
h!0
R 1+h
0
Solutions
R 1 t2
2
et dt
e dt
0
.
2)
h(3 + h
Solution First, using that the limit of a product is the product of limits, we get
lim
h!0
Because
1
3+h2
R 1+h
0
R 1 t2
R 1+h t2
R 1 t2
2
et dt
e dt
e dt
e dt
1
0
0
K.l
The fundamental theorems of calculus.
Here are the two basic theorems relating integrals and
derivatives. You should know the proofs of these theorems.
First, we need to discuss "onesided" derivatives.
If a function is defined on an interval [a,b], we
The trigonometric functions.
For the present, we shall assume the following theorem
1 concerning existence of the sine and cosine functions. Later
on, when we study power series, we shall prove this theorem.
Theorem 1. There exist two functions sin x and
99 dx where here [x] is de ned
R1 3
p
Problem : Compute 99 (2x 198)2 x
to be the largest integer x.
Solution By the properties of the integral, we know that the above is equal
to
Z 4
p
4
(x)2 x dx.
p
Now, [ x ] takes the value 0 on [0, 1) and the value
18 014 Problem Set 7 Solutions
Total: 12 points
Problem : Given a function g continuous everywhere such that g(1) = 5 and
R1
Rx
g(t)dt = 2, let f (x) = 1 0 (x t)2 g(t)dt. Prove that
2
0
Z x
Z x
0
f (x) = x
g(t)dt
tg(t)dt,
0
00
0
000
then compute f (1) and
1.1
Theorem. Let m and n pg integers,- let :1) 0. Let
h(y) = WW :9; we.
Then h _i_s_ differentiable, and
h'(y) = dyl.
~Eroof. Step L; We first prove the theorem in the case m = l.
Let f(x) = XI1 for X> 0. Then the inverse function to f, denoted
g(y),
N.1
Ilnteérationj
_T_h_e substitution gm
Apostol proves only one version of the substitution rule, the one given in Theorem 1
following. Sometimes the converse is needed; we prove this result in Theorem 2.
Theorem L Assume Ea; f(u) gig g(x) and g(x) a_r_e
18 014 Problem Set 6 Solutions
Total: 24 points
Problem : A water tank has the shape of a right-circular cone with its vertex
down. Its alititude is 10 feet and the radius of the base is 15 feet. Water leaks out
of the bottom at a constant rate of 1 cubic
18 014 Problem Set 8 Solutions
Total: 24 points
Problem : Compute
Z
1
xf 00 (2x)dx
0
00
given that f is continuous for all x, and f (0) = 1, f 0 (0) = 3, f (1) = 5, f 0 (1) = 2,
f (2) = 7, f 0 (2) = 4.
Solution (4 points) Applying integration by parts (th
H.1
The small sgan theorem and the extremevalue theorem.
There are three fundamental theorems concerning a function
that is continuous on a closed interval [a,b]. The first
is the IntermediateValue Theorem, which is stated and proved
on p. 144 of Aposto
J.l
Exercises on derivatives
1.
3.
Define a new derivative by the formula
3 3
D#f(x) = lim (f(X+h) h- (f(X) _
h+0
Assuming that f and g are continuous, and that D#f(x)
and D#g(x) exist, derive formulas for D#(f(x)g(x) and
D#(l/f(x) in terms of D#f(x) and
15.1 Vector Fields
CHAPTER 15
15.1
(page 554)
VECTOR CALCULUS
Vector Fields
(page 554)
An ordinary function assigns a value f (x) t o each point x. A vector field assigns a vector F (x, y) t o each
point (x, y). Think of the vector as going out from the p
13.1 Surfaces and Level Curves
CHAPTER 13
(page 475)
PARTIAL DERIVATIVES
Surfaces and Level Curves
13.1
J
L
The graph of z = f (x, y) is a surface in xyz space. When f is a linear function, the surface is flat a plane).
When f = x2 + y2 the surface is cur
14.1 Double Integrals
(page 526)
CHAPTER 14
MULTIPLE INTEGRALS
Double Integrals
14.1
(page 526)
sR
The most basic double integral has the form $ JR dA or $ dy dx or $ JR dx dy. It is the integral of 1over
the region R in the xy plane. The integral equals
12.1ThePositionVector
CHAPTER 12
12.1
(page4521
MOTION ALONG A CURVE
The Position Vector
This section explains the key vectors that describe motion. They are functions o f t . In other words, the time
is a "parameter." We know more than the path that the
Practice Exam
Problem 1 Evaluate
R
Solutions
t3
p +t dt
1+t2
Solution The problem can be simpli ed as t3 + t = t(t2 + 1). Then by substitution
with u = t2 + 1 and thus du = 2t dt we have
Z p
Z
1
1
1
2 + 1 dt =
t t
u1/2 du = u3/2 + C = (t2 + 1)3/2 + C.
2
3