MATH 119 Practice Final Spring 2017
Instructor: Xiaowei Wang
May 8th, 8:30am-11:30am, 2017
Problem 1. Find all antiderivatives of each of the following functions
Z
1
dx
1.
7x
ln x
+C
7
Z √
2
√ + 2 x dx
2.
x
solution.
√
4
solution. 4 x + x3/2
3
Z
3.
e−x dx
solution. −e−x + C
Z
x−1
4.
dx
x+1
solution. x − 2 ln(x + 1) + C
Z
Problem 2. Given
3
Z
1
f (x)dx = 3,
Z
−1
1
3
g(x)dx = −1.
f (x)dx = 0 and
1
Find
Z
3
(2f (x) − 3g(x))dx.
1.
1
solution.
Z 3
Z
(2f (x) − 3g(x))dx = 2
1
3
Z
f (x)dx − 3
1
g(x)dx = 2 · 3 − 3 · (−1) = 9.
1
1
3
Z
3
2(f (x) − 3)dx.
2.
−1
solution.
Z 3
(2f (x) − 3)dx
Z
=
1
Z
−1
Z
3
f (x)dx − 3
f (x)dx +
2
−1
3
1
dx
−1
2(3 + 0 − 3(3 − (−1)) = −18.
=
Problem 3. Find the area of the region bounded by the following curves
1. y = x3 and y = 2x2 .
solution. The intersection points of the two curves satisfy x3 = 2x2 , i.e.
x2 (x − 2) = 0. So the area is given by
Z
2
3
2
Z
2
|x − 2x |dx =
0
0
2
2
1 4 16
4
2 3 = .
2x − x dx = x − x =
3 0
4 0
12
3
2
3
since for 0 ≤ x ≤ 2 we have x3 ≤ 2x2 .
2. y = ex and y =
1
from x = 1 to x = 2.
x2
solution. Notice that for 1 ≤ x ≤ 2, we have ex > 1 ≥ 1/x2 , this implies
Z
1
2
2
Z 2
x
1
1
1 x
2
x
e − 1 dx =
e
−
dx
=
e
+
=e −e− 2
2
x2 x
x
1
1
Problem 4. Find
∂f ∂f ∂f
,
,
for the following functions
∂x ∂y ∂z
1. f (x, y, z) = xexz + x4 y + y 3
solution. fx = exz + xzexz + 4x3 y and fy = x4 + 3y 2 and fz = x2 exz .
2. f (x, y, z) = zex/y
solution. fx =
z x/y
zx
e
and fy = − 2 ex/y and fz = ex/y .
y
y
Problem 5. Let
f (x, y) = x2 + 4xy + 2y 4
2
1. Find
∂f ∂f
,
.
∂x ∂y
solution. fx = 2x + 4y and fy = 4x + 8y 3
2. Find
∂2f ∂2f ∂2f
,
,
and D(x, y).
∂x2 ∂y 2 ∂x∂y
solution. fxx = 2, fyy = 24y 2 and fxy = 4. Hence
2
D(x, y) = fxx · fyy − fxy
= 48y 2 − 16.
3. Find all points (x, y) where f (x, y) has a possible relative maximum or
minimum.
solution. The critical points of f satisfy the equation fx = fy = 0, that is
2x + 4y = 4x + 8y 3 = 0. From the first equation we obtain x = −2y, and
plug it into the second one. Then we obtain
−8y + 8y 3 = 8y(−1 + y 2 ) = 0.
So the critical points are (x, y) = (0, 0) and (x, y) = (∓2, ±1).
4. Using second derivative test to determine whether the points found in the
previous step is relative maximum or minimum.
solution. Finally, by applying second derivative test, we have
D(0, 0) = −16 < 0, fxx (0, 0) = 2 > 0.
So (0, 0) is a saddle point, i.e. neither relative maximum nor relative
minimum.
And
D(∓2, ±1) = 48 − 16 = 32 > 0, fxx (∓2, ±1) = 2 > 0.
So both (∓2, ±1) are relative minimum by second derivative test.
Problem 6. Four hundred eighty dollars are available to fence in a rectangular
garden. The fencing for the north and south sides of the garden cost 10 dollar
per foot and the fencing for the east and west sides costs 15 per foot. Find the
dimension of the largest possible garden.
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solution. The constraint given by the cost is given by g(x, y) = 10 · 2x + 15 · y −
480 = 0 and we want to maximize the area f (x, y) = xy.
By Lagrange multiplier, we define F (x, y) = f (x, y) + λg(x, y). Then
∂F
= y + 20λ = 0
∂x
∂F
= x + 30λ = 0
∂y
∂F
= 20x + 30y − 480 = 0
∂λ
Then solving x, y in terms of λ and plug these into the third equation, we obtain
−600λ − 600λ − 480 = 0,
i.e. λ = −.4 and x = 12 and y = 8.
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