MATH 151 Engineering Math I, Spring 2014
JD Kim
Week13 Section 5.7, 6.1
Section 5.7 Antiderivatives
Definition Antiderivative
We call F (x) an antiderivative of f (x) if F ′ (x) = f (x).
d 2
(x ) = 2x.
ex) x2 is an antiderivative of 2x, because
dx
Definition
If F is an antiderivative of f , then the most general antiderivative of f is F (x)+C,
where C is any arbitrary constant.
Examples)
f (x) = k
⇒
f (x) = xn , if n 6= −1
f (x) = x−1
f (x) = ex
f (x) = ax
f (x) = cos x
f (x) = sin x
F (x) = kx + C
⇒
⇒
xn+1
+C
n+1
F (x) = ln |x| + C
F (x) = ex + C
⇒
ax
+C
ln a
F (x) = sin x + C
⇒
F (x) =
⇒
⇒
F (x) =
F (x) = − cos x + C
1
f (x) = sec x tan x
f (x) = sec2 x
f (x) = csc2 x
f (x) = csc x cot x
⇒
⇒
F (x) = sec x + C
F (x) = tan x + C
⇒
F (x) = − cot x + C
⇒
F (x) = − csc x + C
1
⇒ F (x) = arctan x + C
x2 + 1
1
f (x) = √
⇒ F (x) = arcsin x + C
1 − x2
f (x) =
Ex1) Find the most general antiderivative
1-1) f (x) = x3 − 4x2 + 17
1-2) f (x) =
1-3) f (x) =
√
3
x2 −
√
x3
x + x2 − 1
x3
1-4) f (x) = ex + √
4
+ 5(1 − x2 )−1/2
2
1−x
2
Ex2) Find f (x) given that f ′′ (x) = 3ex + 4 sin x, f (0) = 1 and f ′ (0) = 2
3
Ex3) Given the graph of f passes through the point (1,6) and the slope of its
tangent line at (x, f (x)) is 2x + 1, find f (2).
Ex4) A particle is moving according to acceleration a(t) = 3t + 8. Find the
position, s(t), of the object at time t if we know s(0) = 1 and v(0) = −2.
4
Ex5) A stone is thrown downward from a 450m tall building at a speed of 5
meters per second. Derive a formula for the distance of the stone above ground level.
Ex6) Suppose the acceleration of an object at time t is given by a(t) = −3j. Find
the position vector function, r(t), if it is known that v(2) = i − j and r(0) = 0.
5
Ex7) Suppose that g is differentiable on (0, ∞), that x2 g ′ (x) = 2x2 + 3x for every
x > 0, and that g(e) = 1. Determine g.
6
Ex8) A car braked with a constant deceleration of 40 feet per second squared,
producing skid marks measuring 160 feet before coming to a stop. How fast was the
car traveling when the brakes were first applied?
7
Section 6.1 Sigma Notation
Definition
If m and n are a non negative integers, with m < n, then
n
X
i=m
ai = am + am+1 + am+2 + · · · + an−1 + an .
We call i the index of the sum, m the lower limit of the sum, and n the upper limit
of the sum.
10
P
ex)
n = 1 + 2 + 3 + · · · + 9 + 10 = 55
n=1
6
P
Ex9)
1
i=1 i + 1
Ex10)
8
P
xk
k=5
Ex11) Write 1 −
1
1
1
1 1
+ −
+
−
in sigma notation.
4 9 16 25 36
8
Theorem
n
n
P
P
ai
cai = c
1.
i=m
i=m
2.
n
P
(ai ± bi ) =
i=m
3.
n
P
n
P
i=m
ai ±
n
P
bi
i=m
1=n
i=1
4.
n
P
(c) = cn
i=1
5.
n
P
i=1
n
P
i=
n(n + 1)
2
n(n + 1)(2n + 1)
6
i=1
2
n
P
n(n + 1)
3
7.
i =
2
i=1
6.
i2 =
9
Ex12) Evaluate
12-1)
50
P
3
i=1
12-2)
25
P
7
i=3
12-3)
100
P
6i
i=1
12-4)
99
P
i=3
1
1
−
i i+1
n 1
P
12-5) lim
n→∞ i=1 n
" #
3
i
+1
n
10
Ex13) If
4
P
i=1
f (i) = 3 and
4
P
g(i) = 13, find
i=1
4
P
(2f (i) + g(i)).
i=1
11