Math 220
November 13
I. Find the derivative of the function.
1. g(x) =
Rx
2. g(x) =
Rx
3. g(x) =
R2
4. f (x) =
R 2x
5. h(t) =
R 2t
6. g(t) =
R t3
1
dt
0 1+t2
t sin(t)dt
2
t2 cos(t2 )dt
x
0
t
t2
cos(t2 + 1)dt
tan(x3 )(x2 + 1)dx
(x + 2)−1 dx
7.
Z
cos(x2 )
g(x) =
sec(x)
sin(t)
dt
t2 + 2
II. Find all numbers c that satisfy the conclusion of the Mean Value Theorem.
1. f (x) = x2 + 2x + 1,
2. f (x) = x2 + 1,
3. f (x) = 1/x,
4. f (x) = sin(x),
[0, 1]
[−1, 1]
[1, 3]
[0, π]
III. Show the equation, x3 + ex = 0, has one solution.
IV. Show the equation, x3 + ax2 + cx + d has at most 3 roots.
V. Sketch the region enclosed by the given curves and find its area.
1
1. y = x2 , y = 4
2. y = 5 − x2 , y = 4
3. y = x2 , y = 4x − x2
4. x = y 2 , x = 4
5. y = x3 , y = x
6. y = |x|, y = x2 − 2
7. y =
√
x, y = x, x = 4
8. y = cos(x), y = 1 − cos(x), 0 ≤ x ≤ π
9. x = y 2 , y = x2
10. x = y 2 − 2y, y = x
2
1
Solutions
I. Find the derivative of the function.
Rx
1. g(x) = 0
Answer:
1
dt
1+t2
g 0 (x) =
1
1 + x2
Rx
2. g(x) = 2 t sin(t)dt
Answer:
g 0 (x) = x sin(x)
R2
3. g(x) = x t2 cos(t2 )dt
Answer:
g 0 (x) = −x2 cos(x2 )
R 2x
4. f (x) = 0 cos(t2 + 1)dt
Answer:
g 0 (x) = cos((2x)2 + 1)2
R 2t
5. h(t) = t tan(x3 )(x2 + 1)dx
Answer:
g 0 (x) = tan((2t)3 )((2t)2 + 1)2 − tan(t3 )(t2 + 1)
R t3
6. g(t) = t2 (x + 2)−1 dx
Answer:
g 0 (x) = (t3 + 2)−1 (3t2 ) − (t2 + 2)−1 2t
7.
Z
cos(x2 )
g(x) =
sec(x)
sin(t)
dt
t2 + 2
Answer:
g 0 (x) =
sin(cos(x2 ))
sin(sec(x))
(− sin(x2 )2x) −
sec(x) tan(x)
2
2
cos (x ) + 2
sec2 (x) + 2
II. Find all numbers c that satisfy the conclusion of the Mean Value Theorem.
3
1. f (x) = x2 + 2x + 1,
Answer:
[0, 1]
4−1
f (1) − f (0)
=
=3
1−0
1
f 0 (c) = 2c + 2
Solve f 0 (c) = 3
2c + 2 = 3
2c = 1
1
c=
2
c=
1
2
is the c satisfy the conclusion of the Mean Value Thm.
2. f (x) = x2 + 1,
Answer:
[−1, 1]
f (1) − f (−1)
2−2
=
=0
1 − −1
2
f 0 (c) = 2c
Solve f 0 (c) = 0
2c = 0
c=0
c = 0 is the c satisfy the conclusion of the Mean Value Thm.
3. f (x) = 1/x,
Answer:
[1, 3]
f (3) − f (1)
1/3 − 1
−1
=
=
3−1
2
3
−1
f 0 (c) = 2
c
1
0
Solve f (c) =
3
−1
−1
=
2
c
3
3 = c2
√
c=± 3
4
c=
√
3 is the c satisfy the conclusion of the Mean Value Thm.
4. f (x) = sin(x),
[0, π]
III. Show the equation, x3 + ex = 0, has one solution.
Answer:
Let f (x) = x3 + ex , f (0) = 1 and f (−2) < −7, thus by the intermediate value theorem
f (x) = 0 for an x ∈ (−2, 0).
Now we show that f (x) has exactly one solution by showing it can’t have two.
Assume that equation has two solution.
Then we have a, b such that f (a) = f (b) = 0. By the Mean value theorem we have there exist
(a)
= 0. f 0 (c) = 3c2 + ec . However, 3c2 + ec ≥ ec > 0. So
c ∈ (a, b) such that f 0 (c) = f (b)−f
b−a
f 0 (c) 6= 0 for all c. We get a contradiction. Thus f (x) cannot have two roots and x3 + ex = 0
has one solution.
IV. Show the equation, x3 + ax2 + cx + d has at most 3 roots.
Let f (x) = x3 + ax2 + cx + d. Assume the f (x) has four roots then there exist x1 , x2 , x3 , x4
such that f (x1 ) = f (x2 ) = f (x3 ) = f (x4 ) = 0. By the mean value theorem have there exists
c1 , c2 , c3 such that f 0 (c1 ) = f 0 (c2 ) = f 0 (c3 ) = 0. However, f 0 (c) = 3x2 + 2ax + c has only two
soluitons since it is a quadratic.
V. Sketch the region enclosed by the given curves and find its area.
1. y = x2 , y = 4
Answer:
5
5
4
3
2
1
0
-3
-2
-1
0
1
2
3
Intersections points: (−2, 4), (2, 4)
The region is between x = −2 and x = 2
The top boundary is y = 4
The bottom boundary y = x2
Z
2
(4 − x2 )dx
Area =
−2
x 3 i2
= 4x −
3 −2
8
8
= (8 − ) − (−8 + )
3
3
16
= 16 −
3
32
=
3
2. y = 5 − x2 , y = 4
Answer:
6
5.0
4.8
4.6
4.4
4.2
4.0
-2
-1
0
1
2
Intersections points: (−1, 4), (1, 4)
The region is between x = −1 and x = 1
The top boundary is y = 5 − x2
The bottom boundary y = 4
Z
1
(5 − x2 − 4)dx
Area =
−1
1
Z
(1 − x2 )dx
=
−1
x3 i 1
3 −1
1
1
= (1 − ) − (−1 + )
3
3
2
=2−
3
4
=
3
=x−
3. y = x2 , y = 4x − x2
Answer:
7
5
4
3
2
1
0
-1
-1
0
1
2
3
Intersections points: (0, 0), (2, 4)
The region is between x = 0 and x = 2
The top boundary is y = 4x − x2
The bottom boundary y = x2
Z
2
(4x − x2 − x2 )dx
Area =
Z0 2
=
(4x − 2x2 )dx
0
2x3 i2
= 2x −
3 0
16
= (8 − ) − (0 − 0)
3
16
=8−
3
8
=
3
2
4. x = y 2 , x = 4
Answer:
8
3
2
1
0
-1
-2
-3
-1
0
1
2
3
4
5
Intersections points: (4, −2), (4, 2)
The region is between y = −2 and y = 2
The right boundary is x = 4
The left boundary x = y 2
Z
2
(4 − y 2 )dy
Area =
−2
y 3 i2
= 4y −
3 −2
8
8
= (8 − ) − (−8 − )
3
3
16
= 16 −
3
32
=
3
5. y = x3 , y = x
Answer:
9
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Intersections points: (0, 0), (1, 1)
The region is between x = 0 and x = 1
The top boundary is y = x
The bottom boundary y = x3
Z
1
Area =
(x − x3 )dx
0
x2 x4 i 1
=
−
2
4 0
1 1
= ( − ) − (0 − 0)
2 4
1
=
4
6. y = |x|, y = x2 − 2
Answer:
10
2
1
0
-1
-2
-2
0
-1
1
2
Intersections points: (2, 2), (−2, 2)
The region is between x = −2 and x = 2
The top boundary is y = |x|
The bottom boundary y = x2 − 2
Z
2
Area =
(|x| − (x2 − 2))dx
−2
0
Z
Z
2
2
(−x − (x − 2))dx +
(x − (x2 − 2))dx
−2
0
Z 0
Z 2
=
(−x − x2 + 2)dx +
(x − x2 + 2)dx
=
−2
0
i0
i2
x
x
x2 x3
= (− −
+ 2x)
+( −
+ 2x)
2
3
2
3
−2
0
8
8
= 0 − (−2 + − 4) + (2 − + 4 − 0)
3
3
8
8
=2− +4+2− +4
3
3
16
= 12 −
3
20
=
3
2
3
√
7. y = x, y = x, x = 4
Answer:
11
2.0
1.5
1.0
0.5
0.0
0
1
2
3
4
Intersections points: (0, 0), (1, 1), (4, 4), (4, 2)
There are two regions, one between x = 0 and x = 1 and the other between x = 1 and
x = 4 For the first region:√
The top boundary is y = x
The bottom boundary y = x
For the second region:
The top boundary is y = x √
The bottom boundary y = x
Z
4
Area =
Z0 1
=
√
| x − x|dx
√
( x − x)dx +
0
Z
4
(x −
1
2
√
x)dx
i4
2
x2 i 1
x
2
= x3/2 −
+ ( − x3/2
3
2 0
2
3
1
2 1
16
1 2
= ( − ) − (0 − 0) + (8 − ) − ( − )
3 2
3
2 3
4
16
= −1+8−
3
3
−12
=
+7
3
9
=
3
=3
12
8. y = cos(x), y = 1 − cos(x), 0 ≤ x ≤ π
Answer:
2
1
0
-1
-2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Intersections points: (π/3, 1/2)
The first region is between x = 0 and x = π/3
The top boundary is y = cos(x)
The bottom boundary y = 1 − cos(x)
The second region is between x = π/3 and x = π
The top boundary is y = 1 − cos(x)
The bottom boundary y = cos(x)
Z
π
|(1 − cos(x)) − (cos(x))|dx
Area =
0
Z
π/3
Z
π
(cos(x) − (1 − cos(x))dx +
=
0
Z
π/3
Z
π
(2 cos(x) − 1)dx +
=
((1 − cos(x)) − cos(x))dx
π/3
0
((1 − 2 cos(x))dx
π/3
iπ
iπ/3
= 2 sin(x) + x
+ (x − 2 sin(x))
π/3
0
√
π
π √
= 3 + − (0 + 0) + (π + 0) − ( − 3)
3
√3
=π+2 3
9. x = y 2 , y = x2
Answer:
13
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Intersections points: (0, 0), (1, 1)
The region is between x =√0 and x = 1
The top boundary is y = x
The bottom boundary y = x2
Z
Area =
1
√
( x − x2 )dx
0
2
x3 i 1
= ( x3/2 − )
3
3 0
2 1
= − − (0 − 0)
3 3
1
=
3
10. x = y 2 − 2y, y = x
Answer:
14
3.0
2.5
2.0
1.5
1.0
0.5
0.0
-1
0
1
2
3
Intersections points: (0, 0), (3, 3)
The region is between x = 0 and x = 3
The right boundary is y = x
The left boundary y = y 2 − 2y
Z
3
(y − (y 2 − 2y))dx
Area =
Z0 3
=
(−y 2 + 3y)dx
0
y 3 3y 2 i3
= (− +
)
3
2 0
27
− (0 − 0)
= −9 +
3
18
=
3
15