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MA 140 - Review for Final Exam - Spring 2009
- Using the deÖnition of the deÖnite integral, evaluate
R 5
3 (2x (^2) + x 4) dx.
x =
n
n x i = 3 + i
n
2 i n
Z (^5)
3
2 x^2 + x 4
dx = (^) nlim!
X^ n
i=
2 i n
2 i n
n
= lim n!
n
X^ n
i=
12 i n
4 i^2 n^2
2 i n
= (^) nlim!
n
X^ n
i=
24 i n
8 i^2 n^2
2 i n
= (^) nlim!
n
X^ n
i=
26 i n
8 i^2 n^2
= (^) nlim!
n
X^ n
i=
X^ n
i=
26 i n
X^ n
i=
8 i^2 n^2
= (^) nlim!
n
X^ n
i=
n
X^ n
i=
i +
n^2
X^ n
i=
i^2
= (^) nlim!
n
17 n +
n
n (n + 1) 2
n^2
n (n + 1) (2n + 1) 6
= (^) nlim!
n
17 n + 13 (n + 1) +
4 (n + 1) (2n + 1) 3 n
= (^) nlim!
26 (n + 1) n
8 (n + 1) (2n + 1) 3 n^2
- Using n equal subintervals and the right-hand endpoint of each subinterval, use the
deÖnition of the deÖnite integral to compute
R 2
1 (x
(^2) + 1) dx.
x =
n
n
x i = 1 + i n Z (^2)
1
(x^2 + 1) dx = (^) nlim!
X^ n
i=
i n
n
= (^) nlim!
n
X^ n
i=
2 i n
i^2 n^2
= lim n!
n
X^ n
i=
2 i n
i^2 n^2
= lim n!
n
X^ n
i=
X^ n
i=
2 i n
X^ n
i=
i^2 n^2
= lim n!
n
X^ n
i=
n
X^ n
i=
i +
n^2
X^ n
i=
i^2
= (^) nlim!
n
2 n +
n
n (n + 1) 2
n^2
n (n + 1) (2n + 1) 6
= lim n!
(n + 1) n
(n + 1) (2n + 1) 6 n^2
- Evaluate: (^) nlim!
X^ n
i=
2 i n
n
: Also, write a deÖnite integral that this expression is equal to. Here, f (x) = x^3 ; ci = 1 + (^2) ni ; and x = (^) n^2 : This tells us that a = 1 (from ci = a + ix)
(b) We see t = 0: 5 and ci = 0: 5 i: However, in this case i runs from 0 to 9. So,
distance
X^9
i=
v (ci) t
(v (0) + v (0:5) + v (1) + v (1:5) + v (2) + v (2:5) + v (3) + v (3:5) + v (4) + v (4:5)) t = (0 + 4:67 + 7:34 + 8:86 + 9:73 + 10:22 + 10:51 + 10:67 + 10:76 + 10:81) (0:5) = 41 : 785
(c) We see t = 1 and ci = 0:5 + i: In this case i runs from 0 to 4. So,
distance
X^4
i=
v (ci) t
(v (0:5) + v (1:5) + v (2:5) + v (3:5) + v (4:5)) t = (4:67 + 8:86 + 10:22 + 10:67 + 10:81) (1) = 45 : 23
- Evaluate:
R (^) x (^3) 2 x (^2) +x 1 p (^3) x dx Z x^3 2 x^2 + x 1 p (^3) x dx =
Z
x^3 2 x^2 + x 1
x ^1 =^3 dx
=
Z
x^8 =^3 2 x^5 =^3 + x^2 =^3 x ^1 =^3
dx
=
x^11 =^3
x^8 =^3 +
x^5 =^3
x^2 =^3 + C
- Evaluate:
R
sec x(tan x sec x) dx Z sec x(tan x sec x) dx =
Z
sec x tan x sec^2 x
dx = sec x tan x + C
- Evaluate:
R 3
3 p^ x^2 sin^ x cos^2 x+1 dx Z (^) 3
3
x^2 sin x p cos^2 x + 1
dx = 0 since the limits of integration are equal
- Evaluate:
R 1
0 dx 1+x^2 Z (^1)
0
dx 1 + x^2
= tan ^1 x (^10)
= tan ^1 1 tan ^1
- Evaluate:
R (^) x (^3) 4 x (^2) + p (^3) x dx Z x^3 4 x^2 + 5 p (^3) x dx =
Z
x^3 4 x^2 + 5
x ^1 =^3 dx
=
Z
x^8 =^3 4 x^5 =^3 + 5x ^1 =^3
dx
=
x^11 =^3
x^8 =^3 +
x^2 =^3 + C
- If g (x) =
R (^) sec x e^2 x
p x^4 + 5 dx, Önd g^0 (x).
g (x) =
Z (^) sec x
e^2 x
p x^4 + 5 dx
Z 0
e^2 x
p x^4 + 5 dx +
Z (^) sec x
0
p x^4 + 5 dx
Z (^) e 2 x
0
p x^4 + 5 dx +
Z (^) sec x
0
p x^4 + 5 dx
g^0 (x) =
q (e^2 x)^4 + 5 e^2 x^ 2 +
p sec^4 x + 5 sec x tan x
= 2 e^2 x
q (e^2 x)^4 + 5 + sec x tan x
p sec^4 x + 5
- If F (x) =
R (^3) x 2 15 sin
(^4) (t (^2) 7 t) dt; Önd F 0 (x) :
F (x) =
Z (^3) x 2
15
sin^4
t^2 7 t
dt
F 0 (x) = sin^4
3 x^2
3 x^2
6 x = 6 x sin^4
9 x^4 21 x^2
- Let F (x) =
Z (^) tan x
3
p (^5) t (^3) + 7 dx. Find F 0 (x).
F (x) =
Z (^) tan x
3
p (^5) t (^3) + 7 dx
F 0 (x) = 5
p tan^3 x + 7 sec^2 x = sec^2 x 5
p tan^3 x + 7
- The graph of y = f (x) is given below. Use it to compute: (a)
R 3
R 4 ^4 f^ (x)^ dx^ (b) 1 f^ (x)^ dx^ (c)^
R 4
4 f^ (x)^ dx^ (d)^
R 4
4 jf^ (x)j^ dx
(a) Z (^3)
2
f (x) dx =
Z 3
2
x^2 x 2
dx
x^3
x^2 2 x
2
(3)^3
(3)^2 2 (3)
( 2)^3
( 2)^2 2 ( 2)
(b) Here is a graph:
We see f (x) = 0 when x = 1 and x = 2: Since the deÖnite integral counts area under the x-axis as negative, we need to compute three integrals to Önd the area of the region:
Area =
Z 1
2
x^2 x 2
dx
Z 2
1
x^2 x 2
dx +
Z 3
2
x^2 x 2
dx
x^3
x^2 2 x
2
x^3
x^2 2 x
1
x^3
x^2 2 x
2
( 1)^3
( 1)^2 2 ( 1)
( 2)^3
( 2)^2 2 ( 2)
(2)^3
(2)^2 2 (2)
( 1)^3
( 1)^2 2 ( 1)
(3)^3
(3)^2 2 (3)
(2)^3
(2)^2 2 (2)
- Find f (x) if f 00 (x) = 3ex^ + 5 sin x
p x; f (0) = 1; f 0 (0) = 2.
f 0 (x) =
Z
f 00 (x) dx
=
Z
3 ex^ + 5 sin x x^1 =^2
dx
= 3
Z
ex^ dx + 5
Z
sin x dx
Z
x^1 =^2 dx
= 3 ex^ 5 cos x
x^3 =^2 + C
f 0 (0) = 2 = 3e^0 5 cos 0
(0)^3 =^2 + C
C = 4
f (x) =
Z
f 0 (x) dx
=
Z
3 ex^ 5 cos x
x^3 =^2 + 4
dx
= 3
Z
ex^ dx 5
Z
cos x dx
Z
x^3 =^2 dx + 4
Z
dx
= 3 ex^ 5 sin x
x^5 =^2 + 4x + C
f (0) = 1 = 3e^0 5 sin 0 (0)^5 =^2 + 4 (0) + C C = 2 So, f (x) = 3ex^ 5 sin x 154 x^5 =^2 + 4x 2 :
- Find f (x) if f 00 (x) = cos x, f 0 (0) = 1, and f (0) = 2.
f 0 (x) =
Z
f 00 (x) dx
=
Z
cos x dx = sin x + C
f 0 (0) = 1 = sin 0 + C C = 1
f (x) =
Z
f 0 (x) dx
=
Z
(sin x + 1) dx
=
Z
sin x dx +
Z
dx = cos x + x + C
- A rectangular package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 108 inches. Find the dimensions of the package of maximum volume that can be sent. (Assume the cross section is square.) Here is a picture:
x
x y
We want to maximize the function V = x^2 y: This fuinction has two variables; we want only one. Using the given, we see:
4 x + y = 108 y = 108 4 x
Plugging this into the function, we have V (x) = x^2 (108 4 x) : Di§erentiating we have V 0 (x) = 12x (18 x) : We see V 0 (x) is undeÖned nowhere, and V 0 (x) = 0 when x = 0; 18. Since x = 0 makes no sense, we have x = 18 as our only critical number. Using the Second Derivative Test:
V 00 (18) = 216 < 0
So, x = 18 is a maximum. The value of y that corresponds to this value of x is y = 108 4 (18) = 36: Hence, the dimensions of the box of maximum volume are 18 in 18 in 36 in.
- Let f (x) =
x^3 x^2 + 1
. Then, f 0 (x) =
x^2 (x^2 + 3) (x^2 + 1)^2 , and f 00 (x) =
2 x(3 x^2 ) (x^2 + 1)^3
.. Be sure to include the domain, the x-intercept(s) and y-intercept, any horizontal and vertical asymptotes, intervals of increase and decrease, any relative extrema, intervals of posi- tive and negative concavity, and points of ináection. Using this information, sketch a graph of f (x): Please note the typo in the second derivative. It should be x^2 in the numerator, not x^3. The domain of f (x) is ( 1; 1 ) : The x-intercept is (0; 0) (set the top equal to zero and solve). The y-intercept is (0; 0) (plug 0 in for x and solve for y).
There are no horizontal or vertical asymptotes. We see f 0 (x) is undeÖned nowhere, and f 0 (x) = 0 when x = 0: (Note that x^2 + 3 is never 0.) Using a sign chart:
f'(x)
So, f is increasing on ( 1; 1 ) (or ( 1; 0)[(0; 1 )). Thus, f has no realtive extrema. We see f 00 (x) is undeÖned nowhere, and f 00 (x) = 0 when x = 0;
p 3 : Using a sign chart:
f''(x)
So, f is concave up on
p 3
[ (0; 3) and concave down on
p 3 ; 0
[
p 3 ; 1
So, f has ináection points at
p 3 ; f
p 3
p 3 ; ^34
p 3
; (0; f (0)) = (0; 0) ; and
p 3 ; (^34)
p 3
Here is a graph:
- An open box with a square base is to be constructed from a piece of cardboard 12 inches on a side by cutting out a square from each corner and turning up the sides. Express the volume V of the box as a function of the length of the side of the square cut from each corner. Find the length of the side of the square cut out that maximizes the volume of the box. What is the maximum volume?
x x
x
x
x x
x
x
(^) 12 inches - 6
?
12 inches
Here is a picture of the box:
Since
x^ lim! 0
5 ln (cos 3x) x = (^) xlim! 0
5 (^) cos 3^1 x ( sin 3x) 3 1 = (^) xlim! 0 ( 15 tan 3x) = 0 we see: lim x! 0 (cos 3x)^5 =x^ = e^0 = 1
(b)
x^ lim! 0
x csc x
= lim x! 0
x
sin x
= lim x! 0
sin x x x sin x
= lim x! 0
cos x 1 sin x + x cos x
= lim x! 0
sin x cos x + cos x x sin x
(c)
lim x! 0 +^ x^2 cot x = lim x! 0 +
x^2 tan x = lim x! 0 +
2 x sec^2 x =
(d)
lim x! 2
x^2 4
x x 2
= lim x! 2
8 x (x + 2) x^2 4
= lim x! 2
x^2 2 x + 8 x^2 4 = lim x! 2
2 x 2 2 x =
(e)
x^ lim! 0
sin^2 x
x^2
= lim x! 0
x^2 sin^2 x x^2 sin^2 x
= lim x! 0
2 x 2 sin x cos x 2 x sin^2 x + 2x^2 sin x cos x
= lim x! 0
2 2 cos^2 x + 2 sin^2 x 2 sin^2 x + 8x sin x cos x + 2x^2 cos^2 x 2 x^2 sin^2 x
= lim x! 0
8 sin x cos x 12 sin x cos x + 12x cos^2 x 12 x sin^2 x 8 x^2 sin x cos x
= lim x! 0
8 cos^2 x 8 sin^2 x 24 cos^2 x 64 x cos x sin x 24 sin^2 x 8 x^2 cos^2 x + 8x^2 sin^2 x
(f)
t^ lim!
2 t 3 2 t + 5
t = (^) tlim!1 eln(
22 tt +5 (^3) )t
= (^) tlim!1 et^ ln( 22 tt +5 (^3) )
= (^) tlim!1 e
ln
2 t 3 2 t+
t ^1
= e tlim!
ln
2 t 3 2 t+
t ^1
Since
t^ lim!
ln
2 t 3 2 t+
t ^1 = (^) tlim! ln (2t 3) ln (2t + 5) t ^1 = (^) tlim!
2 2 t 3 ^
2 2 t+ t ^2 = (^) tlim!
2 (2t + 5) 2 (2t 3) (2t 3) (2t + 5)
t^2
= lim t!
16 t^2 4 t^2 + 4t 15
= (^) tlim!
32 t 8 t + 4
= (^) tlim!
We see f 0 (x) is undeÖned nowhere (since x^2 + 3 is never zero), and f 0 (x) = 0 when x = 0: (Note that x^2 + 3 is never 0.) Using a sign chart:
f'(x)
So, f is increasing on ( 1; 0) and decreasing on (0; 1 ). So, f has a realtive maximum at (0; f (0)) = (0; 0) We see f 00 (x) is undeÖned nowhere, and f 00 (x) = 0 when x = 1 : Using a sign chart:
f''(x)
So, f is concave up on ( 1; 1) [ (1; 1 ) and concave down on ( 1 ; 1) : So, f has ináection points at ( 1 ; f ( 1)) =
; and
- Find the absolute maximum and absolute minimum of the function f (x) = x^2 x 2
on the interval [3; 7].
f 0 (x) = 2 x (x 2) x^2 (1) (x 2)^2 = x^2 4 x (x 2)^2
We see f 0 (x) is undeÖned at x = 2, and f 0 (x) = 0 when
x^2 4 x = 0 x (x 4) = 0 x = 0 ; 4
However, 0 and 2 are not in [3; 7] : So, we do not consider them. Checking the critical numbers that are in the interval as well as the endpoints, we see:
x f (x) 3 9 4 8 - absolute minimum 7 495 - absolute maximum
- Sketch a graph of a function that satisÖes all of the given conditions: (a) f (0) = 0 (b) f 0 ( 2) = f 0 (1) = f 0 (9) = 0 (c) (^) xlim!1 f (x) = 0 (d) (^) xlim! 6 f (x) = 1 (e) f 0 (x) < 0
on ( 1; 2), (1; 6), and (9; 1 ) (f) f 0 (x) > 0 on ( 2 ; 1) and (6; 9) (g) f 00 (x) > 0 on ( 1; 0) and (12; 1 ) (h) f 00 (x) < 0 on (0; 6) and (6; 12):
Condition (a) gives that the point (0; 0) is on the curve. Condition (b) says the function áattens out at x = 2 ; 1 ; 9 : Using conditions (e) and (f) with this information, we can make a sign chart:
f'(x)
So, f is increasing on ( 2 ; 1) and (6; 9) and decreasing on ( 1; 2), (1; 6), and (9; 1 ): We see f (x) has a relative maximum at x = 1 and x = 9; and f (x) has a relative minumum at x = 2.
Condition (c) gives that f (x) has a horizontal asymptote of the x-axis. Condition (d) gives that f (x) has a vertical asymptote of x = 6. Further, we know that on both sides of x = 6; f (x) approaches 1.
Conditions (g) and (h) give the following sign chart:
f''(x)
So, f is concave up on ( 1; 0) [ (12; 1 ) and concave down on (0; 6) [ (6; 12) : So, f has ináection points at x = 0 and x = 12: Here is a graph:
We see f 0 (x) is undeÖned nowhere, and f 0 (x) = 0 when x = 73 ; 3 : However, 73 is not in [ 4 ; 2] : So, we do not consider it. Checking the critical numbers that are in the interval as well as the endpoints, we see: x f (x) 4 34 3 43 - absolute maximum 2 32 - absolute minimum
- Find the critical numbers of the function F (x) = x^4 =^5 (x 4)^2.
F 0 (x) =
x ^1 =^5 (x 4)^2 + x^4 =^5 (2 (x 4))
= x ^1 =^5 (x 4)
(x 4) + 2x
= x ^1 =^5 (x 4)
x
We see F 0 (x) is undeÖned when x = 0; and F 0 (x) = 0 when:
x ^1 =^5 (x 4)
x
x 4 = 0 OR
x
x = 4 OR x =
So, the critical numbers are 0 ; 4 ; and 87.
- Use a linearization or di§erentials to approximate
p 24 : 1. Let f (x) = p x = x^1 =^2 ; x = 25; dx = 0 : 9 : Then:
dy = f 0 (x) dx dy =
x ^1 =^2 dx
dyj (^) x= dx= 0 : 9
(25) ^1 =^2 ( 0 :9)
So:
f (24:1) f (25) dy f (24:1) 5 0 : 09 = 4 : 91
- Answer the following questions about the proof of the Mean Value Theorem:
(a) What does the equation y =
f (b) f (a) b a
(x a) + f (a) represent? This is the equation of the secant line through the points (a; f (a)) and (b; f (b)) : (b) We deÖne g(x) = f (x) y. Explain why g(a) = 0 = g(b). The function g (x) gives the distance between the function and the secant line. Since the function and the secant line both pass through the point (a; f (a)) and (b; f (b)) ; the distance between the function and the secant line at x = a and x = b is 0. (c) What theorem do we apply to g(x)? Why is that theorem applicable here? We apply Rolleís Theorem to g (x) : We can apply Rolleís Theorem to g (x) because g (x) is the di§erence of two continuous and di§erentiable functions, f (x) and y (y is a line); making g (x) continuous and di§erentiable as well.
- The graph below is the derivative f 0 (x) of a function f (x).
(a) On what intervals is f (x) increasing? decreasing? The function f (x) is increasing when f 0 (x) is positive. From the above graph, we see that f 0 (x) is positive on the intervals ( 2 ; 0) and (4; 1 ) : So, f (x) is increasing on ( 2 ; 0) [ (4; 1 ) : The function f (x) is decreasing when f 0 (x) is negative. From the above graph, we see that f 0 (x) is negative on the intervals ( 1; 2) ; (0; 2) and (2; 4) : So, f (x) is decreasing on ( 1; 2) [ (0; 2) [ (2; 4) : (b) At what x values does f (x) have a local maximum? local minimum? The function f (x) will have a local maximum when f 0 (x) chances from positive to negative, or when f (x) changes from increasing to decreasing. We see that this happens at x = 0:
