(ang684) griffin HW04 Gilbert (56650) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Thus we need to find
1 0
1
f (x)dx = F (1) - F (0) .
Now
1 1
The graph of f is shown in the figure
0
f (x) dx =
0
x dx =
1 2 x 2
1 0
=
1 . 2
Consequently, 6 4 2 F (8) - F (0) = 31 1 + = 16 . 2 2
keywords: velocity, distance, graph analysis, fundamental theorem 2 4 6 8 002 10.0 points
If F is an anti-derivative of f and
8 1
Evaluate the definite integral
6
31 , f (x)dx = 2
I =
0
3 sin 2x - 2 cos2 x dx . cos x 3 2
find the value of F (8) - F (0). 35 1. F (8) - F (0) = 2 2. F (8) - F (0) = 18 3. F (8) - F (0) = 17 33 4. F (8) - F (0) = 2 5. F (8) - F (0) = 16 correct Explanation: We already know that the area under the 31 graph on the interval 1 x 8 is equal to , 2 alternatively, by the Fundamental Theorem of Calculus we can say that 31 F (8) - F (1) = . 2 On the other hand,
8 1 8
1. I = 6 - 2. I = 3 -
3. I = 7 - 3 3 4. I = 6 + 3
5. I = 5 - 3 3 correct 6. I = 6 + 2
Explanation: Since sin 2x = 2 sin x cos x , the integrand can be rewritten as 6 sin x cos x - 2 cos2 x cos x = 2(3 sin x - cos x) .
f (x)dx =
0 0
f (x)dx +
1
f (x)dx.
griffin (ang684) HW04 Gilbert (56650) Thus
6
2
I= 2
0
(3 sin x - cos x) dx
6
1. I = 2 3 - 1 2. I = -2 3. I = -1 4. I = 2 3 + 2 5. I = 2 3 + 1 6. I = -2 7. I = 2 3 - 2 correct 8. I = 1
= 2 - 3 cos x - sin x = 2 - Consequently, I = 5-3 3 . 003
4
0
3 1 3- 2 2
+ 6.
10.0 points
Evaluate the definite integral I =
1
4 x 3+ x
dx .
Explanation: Since 1 = sec2 , 2 cos we see that
/6
1. I = 22 correct 2. I = 23 3. I = 26 4. I = 24 5. I = 25 Explanation: We first expand x 3+ 4 x 4 = 3 x+ , x
4 1
d tan = sec2 , d
I =
0
3 sec2 - 2 sin d
/6 0
=
3 tan + 2 cos =
3 + 3 - 2. 3
Consequently, I = 2 3-2 005 .
and then integrate term by term. This gives I = Consequently, I = 22 004 . I = 10.0 points 1. I = 2. I = 79 6 25 2 2x3/2 + 8x1/2 .
10.0 points
Find the value of the integral
5 0
4x - x2 dx .
Evaluate the integral
/6
I =
0
3 - 2 sin d . cos2
griffin (ang684) HW04 Gilbert (56650) 3. I = 13 correct 4. I = 5. I = 38 3 77 6
3
How far above the ground will the helicopter be after 3 seconds? 1. height = 69 feet 2. height = 63 feet correct 3. height = 71 feet 4. height = 67 feet
2
Explanation: The graph of f (x) = 4x - x is a parabola
5. height = 65 feet Explanation: The height h of the helicopter after 3 seconds is given by the definite integral
3
4
5
h =
0
12t + =
30 (t + 2)2 30 t+2
3 0
dt .
(not drawn to scale) which lies above the xaxis on [0, 4] and below on [4, 5]. Thus |4x - x | = So
4 5 2
6t2 -
Consequently, height = 63 feet. 007 10.0 points
4x - x2 ,
x2 - 4x,
0 x 4,
4 x 5.
Evaluate the definite integral
0
I = =
(4x - x2 ) dx + 1 2x - x3 3
2 4
4
(x2 - 4x) dx 1 3 x - 2x2 3
5
/3
I =
/6
(4 sin 2x + 2 cos 2x) dx .
+
0
.
4
Consequently, I = 13 . 006 10.0 points
5 3 2 2. I = 3 1. I = 3. I = 3 4. I = 2 correct 5. I = 2 3 6. I = 4 Explanation: To reduce the integral to one involving just sin u and cos u, set u = 2x.
A helicopter rises vertically and t seconds after leaving the ground its velocity is given in feet per second by 30 . v(t) = 12t + (t + 2)2
griffin (ang684) HW04 Gilbert (56650) Then du = 2 dx, so I = 1 2
2/3
4
To evaluate the integral set u = 2x2 + 1. For then du = 4x dx, in which case f (x) = . 1 2 = u1/2 du = 1 3/2 u +C 3
(4 sin u + 2 cos u) du
/3
= But
1 -4 cos u + 2 sin u 2
2/3 /3
1 (2x2 + 1)3/2 + C , 3
where C has to be chosen so that 1 2 = - , cos 3 2 2 sin = 3 3 , 2 f (2) = 5 , Thus f (x) = 1 (2x2 + 1)3/2 - 27 3 + 5. i .e., C + 9 = 5.
while cos 1 = , 3 2 sin = 3 3 . 2
Consequently, the graph has y-intercept = - 11 3 .
Consequently, I = 2 .
009
10.0 points
008
10.0 points
Evaluate the definite integral
4
The graph of f has slope df = 2x 2x2 + 1 dx and passes through the point (2, 5). Find the y-intercept of this graph. 1. y-intercept = - 2. y-intercept = - 13 3 11 correct 3
I =
1
3 dx . x( x + 3)2
1. I = 2. I = 3. I = 4. I = 5. I =
9 20 2 5 7 20 1 4 3 correct 10
3. y-intercept = -3 4. y-intercept = -4 5. y-intercept = - 10 3
Explanation: Set u2 = x. Then 2u du = dx, while x = 1 x = 4 In this case,
2
= =
u = 1 u = 2.
Explanation: The function f satisfies the equations f (x) = 2x 2x2 + 1 dx, f (2) = 5.
I = 6
1
1 1 du = 6 - 2 (u + 3) u+3
2 1
.
griffin (ang684) HW04 Gilbert (56650) Thus I = -6 010 1 1 - 5 4 = 3 . 10 2. I = 1 1 +C + 5 sin 2 1 1 +C - 5 cos 2
5
10.0 points
3. I = - 4. I = 5. I =
Determine the indefinite integral I = cos8 x sin x dx .
1 1 + C correct - 5 sin 2 1 1 +C - 5 cos 2 1 1 +C - 5 sin 2
1 1. I = - cos8 x + C 8 1 2. I = - cos7 x + C 7 3. I = 4. I = 1 8 sin x + C 8 1 9 sin x + C 9
6. I = -
Explanation: Set u = 1/. Then du = - so I = - (5 cos u - 2u) du = -5 sin u + u2 + C with C an arbitrary constant. Consequently, I = 1 1 +C . - 5 sin 2 10.0 points 1 d , 2
1 5. I = - cos9 x + C correct 9 6. I = 1 7 sin x + C 7
Explanation: Set u = cos x. Then du = - sin dx , in which case I = - 1 u8 du = - u9 + C 9
012
Evaluate the definite integral
1
with C an arbitrary constant. Consequently, 1 I = - cos9 x + C . 9 011 10.0 points 2. I = d 3. I =
I =
0
8xn-1 dx (8 + 5xn )2
when n is an arbitrary integer, n 2. 1. I = 13 n 8 n 1 correct 13 n 1 n
Determine the indefinite integral I = 1. I = 1 2 5 cos 2 1 -
1 1 +C + 5 cos 2
4. I = -
griffin (ang684) HW04 Gilbert (56650) 5. I = - 5 n 4. Area = 26 sq.units 5. Area = 29 sq.units
6
Explanation: The integral can be evaluated by substitution. Set u = 8 + 5xn . Then du = 5nxn-1 , dx so 8 I = 5n Consequently, I = 8 5n 013 1 1 - 8 13 = 1 . 13n
13 8
Explanation: The graph of f is a parabola opening downwards and crossing the x-axis at x = 0 and x = 3. Thus the required area is similar to the shaded region in the figure below.
8 1 1 du = - 2 u 5n u
13
.
8
10.0 graph points of f In terms of definite integrals, therefore, the required area is given by
3 6
Evaluate the definite integral
12
I =
4
x-7 dx . x-3
Correct answer: 1.33333. Explanation: Set u = x - 3. Then du = dx and
9 0
(3x - x2 ) dx -
3
(3x - x2 ) dx .
I =
1
u-4 du = u
9 1
u1/2 - 4u-1/2 du .
Now
3 0
Consequently, I = 2 3/2 u - 8u1/2 3 014
9 1
(3x - x2 ) dx =
3 2 1 3 x - x 2 3
3 0
=
9 , 2
=
4 . 3
while
6 3
10.0 points
(3x - x2 ) dx =
3 2 1 3 x - x 2 3
6 3
= -
45 . 2
Find the area between the graph of f and the x-axis on the interval [0, 6] when f (x) = 3x - x2 . 1. Area = 30 sq.units
Consequently, Area = 27 sq.units .
keywords: integral, graph, area 2. Area = 28 sq.units 015 3. Area = 27 sq.units correct 10.0 points
griffin (ang684) HW04 Gilbert (56650) Find the area enclosed by the graphs of f (x) = sin x , on [0, /2]. 1. area = 2( 2 + 1) 2. area = 2 + 1 3. area = 4( 2 - 1) 4. area = 4( 2 + 1) 5. area = 2( 2 - 1) correct 6. area = 2 - 1 g(x) = cos x
7
But by the Fundamental Theorem of Calculus, A1 = while A2 = sin + cos
/2 /4
sin + cos
/4 0
=
2- 1,
= 1-
2.
Consequently, area = A1 - A2 = 2( 2 - 1) . 016 10.0 points
Explanation: The area between the graphs of y = f (x) and y = g(x) on the interval [a, b] is expressed by the integral
b
Find the area of the bounded region enclosed by the parabola y = 6 + 4x - x2 and the straight line y = 6x + 3. Correct answer: 10.6667. Explanation: The parabola and the straight line intersect when 6x + 3 = 6 + 4x - x2 , i.e., when x2 + 2x - 3 = (x + 3)(x - 1) = 0. As the parabola opens downwards, the area of the region enclosed by the parabola and the straight line is thus given by
1
A =
a
|f (x) - g(x)| dx ,
which for the given functions is the integral
/2
A =
0
|sin x - cos x| dx .
But, as the graphs y 1
/4 cos : /2 sin :
of y = cos x and y = sin x on [0, /2] show, cos - sin Thus
/4
0,
on [0, /4], on [/4, /2].
-3
0,
(3 - 2x - x2 ) dx = 1 3x - x2 - x3 3
1
.
-3
A =
0 /2
{cos - sin } d {cos - sin } d = A1 - A2 .
Consequently, Area = 32 10.6667 sq. units. 3
-
/4
griffin (ang684) HW04 Gilbert (56650) Find the area of the shaded region in 017 10.0 points Find the area of the bounded region enclosed by the graphs of f (x) = x - 5x, Correct answer: 62.5. Explanation: The graph of f is a parabola opening upwards and crossing the x-axis at x = 0 and 5, while the graph of g is a parabola opening downwards and crossing the x-axis at x = 0 and 5. Thus the required area is the shaded region in the figure below graph of g x
2
8
y P
g(x) = 10x - 2x .
2
bounded by the graph of f (x) = 4x2 , the tangent line at P (2, f (2)), and the x-axis. 1. area = 8 2. area = 2 3. area = 4. area = 16 3 8 correct 3
graph of f In terms of definite integrals, therefore, the required area is given by
5
5. area = 4 Explanation: To find the area of the shaded region it is simplest to compute separately the areas y P
Area =
0
(g(x) - f (x)) dx
5
=
0
(15x - 3x2 ) dx. 15 2 x - x3 2
Now
5 0
(15x - 3x2 ) dx =
5 0
=
125 2 .
Thus Area = 018 125 sq.units. 2 10.0 points x0 2
x
below the parabola and below the tangent line, then subtract.
griffin (ang684) HW04 Gilbert (56650) The area below the parabola is given by the definite integral
2
9
Find the area of this shaded region. 1. area = 2. area = 3. area = 4. area = 5. area = 127 sq. units 6 133 sq. units 6 131 sq. units 6 125 sq. units correct 6 43 sq. units 2
A1 =
0
4x2 dx =
4 3 x 3
2 0
=
32 . 3
On the other hand, the triangular region below the tangent line has area A2 = 1 base length height 2 1 = (2 - x0 ) f (2) , 2
where x0 is the x-intercept of the tangent line. Now f (2) = 16 , f (x) = 8x ,
f (2) = 16 ,
Explanation: The area of the enclosed region is given by the integral
b
so by the point slope formula an equation for the tangent line at P (2, f (2)) is y - 16 = 16(x - 2) , Thus x0 = 1 , and so A2 = 1 1 16 = 8 . 2 i .e., y = 16x - 16 .
I =
a
{g(y) - f (y)} dy
where a, b are the y-coordinates of the points of intersection of the two graphs. Now a, b are the solutions of the equation f (y) = g(y), i.e., y 2 - y - 6 = (y + 2)(y - 3) = 0 . Thus
3
Consequently, the shaded region has area = A1 - A2 = 019 y 10.0 points 8 . 3
I =
-2
y + 6 - y 2 dy = 1 2 1 y + 6y - y 3 2 3
3
The shaded region in Consequently, x
.
-2
area =
125 sq. units . 6 10.0 points
020
is enclosed by the graphs of f (y) = y 2 , g(y) = y + 6 .
A car heads north from Austin on IH 35. Its velocity t hours after leaving Austin is given in miles per hour by 2 v(t) = 10 + t - t2 . 3
griffin (ang684) HW04 Gilbert (56650) Determine how many hours will elapse before the car is next in Austin. 1. car never returns to Austin 2. 7 hours later 3. 8 hours later 4. 5 hours later 5. 6 hours later correct Explanation: Since the car leaves Austin at time t = 0, the position of the car t hours later is the anti-derivative s(t) = of v(t). But 2 (10 + t - t2 ) dt 3 1 1 = 10t + t2 - t3 + C 3 3 1 = t(t + 5) 2 - t + C . 3 Thus 1 s(t) = t(t + 5) 2 - t . 3 Since the car will be in Austin whenever s(t) = 0, we see that after departing at t = 0, the next time it will in Austin will be 6 hours later . Now
0
10 y
x
bounded by the graphs of f (x) = x3 - x2 - 5x, Correct answer: 21.0833. Explanation: The graph of f is a cubic, while the graph of g is a straight line. On the other hand, f (x) - g(x) = x3 - x2 - 6x so the graphs intersect at x = -2, 0, and 3. Thus the required area is given by
0
g(x) = x .
2 (10 + t - t2 ) dt, 3
s(0) = 0
= x(x - 3)(x + 2) ,
Area =
-2
(f (x) - g(x)) dx
3
+
0
(g(x) - f (x)) dx = A1 + A2 .
A1 =
-2
(x3 - x2 - 6x) dx
0
1 4 1 3 x - x - 3x2 4 3 On the other hand, =
3
=
-2
16 . 3
A2 =
0
(-x3 - x2 + 6x) dx =
keywords: word problem, integral, polynomial, distance, time taken, 021 10.0 points
3 63 1 1 . - x4 + x3 + 3x2 = 4 3 4 0 Consequently,the shaded region has
area =
Find the area of the shaded region in
16 63 253 + = sq.units. . 3 4 12