Solutions Cal1 Final - Calculus 1 | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM

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Môn: Calculus 1 (MA001IU)

Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

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Tác giả:

Profile Mai Nguyệt
CALCULUS 1
Solutions for Final Examination
Question 1.
f
(x x) =
2
e
−2x
.
(a)
f
0
(x) = 2xe xe .
−2x
− 2x
2
e
−2x
= 2
−2x
(1 − x)
The increasing/decresing intervals are given in the following chart.
x 0 1
f
0
(x) - 0 + 0 -
f(x) & % &
(b) According to part (a), has local maximum and local minimum value
f f(1) = e
−2
f(0) = 0.
(c)
f
0
(x) = 2(x − x
2
)e
−2x
.. So
f
”(x x x x) = 2(1 − 2 )e
−2x
− 4(x −
2
)e
−2x
= 2(1 − 4x + 2
2
)e
−2x
,
which has the same sign as
p(x) = 1−4 +2 ( 2 1+ 2 2,x x
2
. Since p x) = 0 iff x = 1−
√
/2 < x <
√
/
and
p( 2 2x) < 0 iff 1 −
√
/2 < x < 1 +
√
/2, the f(x) is concave upward in the intervals
x <
1−
√
2/ /2 and x > 1+ 2
√
2, and concave downward in the interval 1−
√
2 1+ 2 2./2 < x <
√
/
The graph of
f has two inflection points: (1 ±
√
2 2 (3 2) )./ , /2 ±
√
e
−2∓
√
2
(d) Sketch the graph of , see Figure 1.f
Figure 1: Graph of in Question 1
f(x) = x
2
e
−2x
Question 2.
a) Let y =
1 − sin
1
x
2x
. Then
lim
x
→∞
ln y = lim
x
→∞
2x ln
1 − sin
1
x
= 2 lim
x→∞
ln
1 − sin
1
x
1/x
= 2 lim
x→∞
( cos(1 )( 1 sin(1 ))
− /x − /x
2
)/(1 − /x
−1/x
2
= 2 lim
x
→∞
−cos(1 )/x
1
− sin(1/x)
= −2.
Thus, lim
x→∞
y = 1 ./e
2
1
b) Let f(x) = 2x − −1 sin(x). We use Newton’s method with starting point = 1 to findx
0
the root of f (x) = 0.
x x
n+1
=
n
−
f(x
n
)
f
0
(x
n
)
= x
n
−
2x
n
− −1 sin(x
n
)
2
− cos(x
n
)
,
which gives us
x
1
= 0. .891395995328754, x , x
2
= 0.887865749400352
3
= 0 887862211574412, x
4
= 0.887862211570866
Question 3. Evaluate the integrals
a)
I
=
Z
2
1
x x
2
ln(
3
)dx =
Z
2
1
ln
xdx
3
=
x
3
ln x
2
1
−
Z
2
1
x
3
(1/x dx)
= 8 ln 2
−
Z
2
1
x x
2
dx = 8 ln 2 −
3
/3
2
1
= 8 ln 2 − 7/3 = 3. .2118
b) Set
u = ln x, dv =
1
√
x
dx. Then, du = dx/x, v = 2
√
x. Thus
Z
e
1
ln x
√
x
dx = (ln x)2
√
x
e
1
−
Z
e
1
2
√
x
x
dx
= 2
√
e −
Z
e
1
2
√
x
dx
= 2
√
e − 4
√
x
e
1
= −2
√
e + 4 = 0 7026.
Question 4.
Z
∞
1
ln x
x
2
dx =
Z
∞
1
ln
xd
−1
x
=
−ln x
x
∞
1
+
Z
∞
1
1
x
2
dx
= 0
−
1
x
∞
1
= 1.
(b)
x
+ e
−x
x
2
>
1
x
, x ≥ 1
The integral
R
∞
1
(1 )/x dx is divergent, therefore, the given integral is also divergent, by Com-
parison Test.
Question 5.
(a) The equation 2 + 1) = 0 gives intersection points− −x (x
2
−2 )(x) = (2 −x x
at x = −1, x = 2. Area:
A
=
Z
2
−1
(2
− x − (x
2
− 2 ))x dx =
Z
2
−1
(2 +
x − x
2
)dx
= (2
x + x
2
/2 − x
3
/3)
2
−
1
= 9/2
2
Figure 2: Area in Question 5, (a)
(b) y = sin( = 4. Radius: = 4x), 0 ≤ ≤x π, and y = 0 about the line x R −x, circumference
2π(4 − x), and height sin x. So,
V
=
Z
π
0
2 8
π(4 − x) sin xdx = − π cos x
π
0
+
Z
π
0
2πxd cos x
= 16
π + 2π(x cos x
π
0
−
Z
π
0
cos )xdx
= 16
π + 2π(−π − sin x
π
0
) = 16π −2π
2
= 30 5263.
3
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CALCULUS 1
Solutions for Final Examination Question 1. f (x) = x2e−2x. (a)
f 0(x) = 2xe−2x − 2x2e−2x = 2xe−2x(1 − x).
The increasing/decresing intervals are given in the following chart. x 0 1 f 0(x) - 0 + 0 - f (x) & % &
(b) According to part (a), f has local maximum f (1) = e−2 and local minimum value f (0) = 0.
(c) f 0(x) = 2(x − x2)e−2x.. So
f ”(x) = 2(1 − 2x)e−2x − 4(x − x2)e−2x = 2(1 − 4x + 2x2)e−2x, √ √
which has the same sign as p(x) = 1−4x+2x2. Since p(x) = 0 iff x = 1− 2/2 < x < 1+ 2/2, √ √
and p(x) < 0 iff 1 − 2/2 < x < 1 + 2/2, the f(x) is concave upward in the intervals √ √ √ √
x < 1− 2/2 and x > 1+ 2/2, and concave downward in the interval 1− 2/2 < x < 1+ 2/2. √ √ √
The graph of f has two inflection points: (1 ± 2/2, (3/2 ± 2)e−2∓ 2).
(d) Sketch the graph of f , see Figure 1.
Figure 1: Graph of f (x) = x2e−2x in Question 1
Question 2. a) Let y = 1 − sin 1 2x. Then x  1  lim ln y = lim 2x ln 1 − sin x→∞ x→∞ x ln 1 − sin 1  = 2 lim x x→∞ 1/x
(− cos(1/x)(−1/x2)/(1 − sin(1/x)) = 2 lim x→∞ −1/x2 − cos(1/x) = 2 lim = −2. x→∞ 1 − sin(1/x) Thus, lim y = 1/e2. x→∞ 1
b) Let f (x) = 2x − 1 − sin(x). We use Newton’s method with starting point x0 = 1 to find the root of f (x) = 0. f (x ) 2x − 1 − sin(x ) x − n = x − n n , n+1 = xn f n 0(x ) 2 ) n − cos(xn which gives us
x1 = 0.891395995328754, x2 = 0.887865749400352, x3 = 0.887862211574412, x4 = 0.887862211570866
Question 3. Evaluate the integrals a) Z 2 Z 2 I = x2 ln(x3)dx = ln xdx3 1 1  2 Z 2 = x3 ln x − x3(1/x)dx 1 1 Z 2 2 = 8 ln 2 −
x2dx = 8 ln 2 − x3/3 = 8 ln 2 − 7/3 = 3.2118.  1 1 √ b) Set u = ln x, dv = 1 √ dx. Then, du = dx/x, v = 2 x. Thus x √ Z e ln x √ e Z e 2 x √ dx = (ln x)2 x − dx  x 1 x 1 1 √ Z e 2 = 2 e − √ dx x 1 √ √ e √
= 2 e − 4 x = −2 e + 4 = 0.7026 1 Question 4. Z ∞ ln x Z ∞ −1 dx = ln xd x 1 x2 1 − ln x∞ Z ∞ 1 =  + dx x  1 x2 1 1 ∞ = 0 −  = 1. x 1 (b) x + e−x 1 > , x ≥ 1 x2 x
The integral R ∞(1/x)dx is divergent, therefore, the given integral is also divergent, by Com- 1 parison Test.
Question 5. (a) The equation 2 − x − (x2 − 2x) = (2 − x)(x + 1) = 0 gives intersection points at x = −1, x = 2. Area: Z 2 Z 2 A = (2 − x − (x2 − 2x))dx = (2 + x − x2)dx −1 −1  2 = (2x + x2/2 − x3/3) = 9/2 −1 2
Figure 2: Area in Question 5, (a)
(b) y = sin(x), 0 ≤ x ≤ π, and y = 0 about the line x = 4. Radius: R = 4−x, circumference
2π(4 − x), and height sin x. So, Z π π Z π V =
2π(4 − x) sin xdx = −8π cos x + 2πxd cos x  0 0 0 π Z π = 16π + 2π(x cos x − cos xdx) 0 0 π
= 16π + 2π(−π − sin x ) = 16π − 2π2 = 30.5263 0 3