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The solutions to the final exam of appm 1350, a mathematics course focusing on calculus, for the summer semester of 2006. It includes answers to various calculus problems, such as determining limits, finding derivatives, and evaluating integrals. Students can use this document as a reference to check their understanding of the concepts covered in the exam.
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(d)
− 5
(ax^2 + bx + c) dx = 2
0
(ax^2 + c) dx (ALWAYS TRUE)
(e) If limx→ 6 f (x)g(x) exists, then the limit is f (6)g(6). (NEITHER)
(a) y = x ln (arccos x)
dy dx =^ (x)
arccos x
1 − x^2
=
−x arccos x
1 − x^2
(b) y = xex
ln y = ex^ ln x ⇒ 1 y
dy dx
= (ex)
x
⇒ dy dx
= y ex
x
= xe
x ex
x + ln^ x
(c) x ey^ = ln xy + arctan y
(x)
ey^ · dy dx
x dy dx
1 + y^2
· dy dx dy dx
x ey^ − 1 y
1 + y^2
x
− ey
dy dx
1 x −^ e
y xey^ − (^1) y − (^) 1+^1 y 2
(a) (^) t→lim 0 + tt
2
lim t→ 0 +^
tt
2 = lim t→ 0 +^
et
(^2) ln t = elimt→^0 +^ t
(^2) ln t
lim t→ 0 +^
t^2 ln t = lim t→ 0 +
ln t t−^2
L =′H lim t→ 0 +
t−^1 − 2 t−^3
= lim t→ 0 +^
t^2 = 0
⇒ lim t→ 0 +^
tt
2 = e^0 = 1
(b) (^) rlim→∞
re^1 /r^ − r
r^ lim→∞
re^1 /r^ − r
= (^) rlim→∞ r
e^1 /r^ − 1
= lim r→∞
e^1 /r^ − 1 1 r L =′H lim r→∞
e^1 /r^ · − r 21 − 1 r^2
= lim r→∞ e^1 /r^ = e^0
= 1
(c) (^) xlim→ 0 x arccot^1 x
x^ lim→ 0 x^ arccot x^1 =^ xlim→ 0
arccot (^1) x 1 x
L =′H lim x→ 0
− 1 1+ (^) x^12 ·^
− 1 x^2 − 1 x^2 = (^) xlim→ 0
1 + (^) x^12
= (^) xlim→ 0
−x^2 x^2 + 1 = 0
(d) (^) hlim→ 0
h
∫ (^) 2+h
2
1 + t^3 dt
h^ lim→ 0
h
∫ (^) 2+h
2
1 + t^3 dt = (^) hlim→ 0
∫ (^) 2+h a
1 + t^3 dt −
a
1 + t^3 dt h = F ′(2) where F (x) =
∫ (^) x
a
1 + t^3 dt
F ′(x) =
1 + t^3 (By FTC pt 1) ⇒ F ′(2) =
0.5 1 1.5^2 2.
1
y=e-x
A
A = (x)(y) = x e−x ⇒ dA dx
= (x)(−e−x) + (1)(e−x) = e−x(1 − x) = 0 ⇒ x = 1 ⇒ A = 1 · e−^1 =^1 e
(b) Plot the function f (x) = x ln x on (0, e]. Find and label all critical points, local and absolute extrema, and inflection points.
f ′(x) = 1 + ln x = 0 ⇒ x =^1 e f ′′(x) = (^1) x > 0 for all x in (0, e]
-1 -0.5 0.5 1 1.5 2 2.5 3
-0.
1
2
3
fHxL=xlnHxL
H ^1 e , - e^1 L
H0,0L
He,eL
(a) What does it mean for f (x) to be continuous at x = a? ⇒ f is continuous at x = a if lim x→a f (x) = f (a).
(b) What does it mean for f (x) to be differentiable at x = a? ⇒ f is differentiable at x = a if lim h→ 0
f (a + h) − f (a) h exists or, alternatively, lim^ x→a
f (x) − f (a) x − a exists.
(c) State both parts of the Fundamental Theorem of Calculus.
FTC Part 1: If f is continuous on [a, b] and F (x) =
∫ (^) x
a
f (t) dt, then for a ≤ x ≤ b
d dx F^ (x) =^
d dx
∫ (^) x
a
f (t) dt = f (x)
FTC Part 2: If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then ∫ (^) b
a
f (x) dx = F (b) − F (a)
(d) If f is a continuous function such that ∫ (^) x
0
f (t) dt = x sin x +
∫ (^) x
0
f (t) 1 + t^2 dt
for all x, find an explicit formula for f (x).
d dx
∫ (^) x
0
f (t) dt = d dx
x sin x +
∫ (^) x
0
f (t) 1 + t^2
dt
f (x) = x cos x + sin x + f^ (x) 1 + x^2 ⇒ f (x) = x^ cos^ x^ + sin^ x 1 − (^) 1+^1 x 2