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Math 50B — Integral Calculus March 17, 2017
Midterm Exam #1 Name: Answer Key David Arnold
Instructions. (100 points) This exam is open notes, open book. This includes any supplementary texts or online documents. You are not allowed to work in groups or pairs on the quiz. You are not allowed to enlist the aid of a tutor or friend to help with the quiz. You must answer all of the exercises on your own. You are also not allowed to share your work with your fellow students.
Honor Pledge: I promise that the solutions submitted were done entirely by me. I received no help from colleagues, friends, or tutors. I also did not share any parts of my solutions with any of my classmates.
Signature:
Please staple this cover and honor pledge atop your solutions.
Neatness Requirement: Each problem on this exam must be done on a separate sheet of paper. All pencil and paper calculations must be done using pencils. Deductions will be made for any work done in pen. Any mistakes made must be carefully erased with an eraser. If there is any scratching out of work or any disorganized presentation, points will be deducted.
( 10 pts) 1. Sketch the region bounded by x = y^2 and y = x − 2. Draw the typical element of area (you may choose a direction of your choice). Label its endpoints with the coordinates, state the area of this typical element, the set up an integral and integrate to find the exact area of this region. Solution: Start with a sketch.
x
y
y = x − 2
x = y^2
(1, −1)
(4, 2)
(y^2 , y) (y + 2, y)
Hence, the area of the shaded rectangle is: dA =
[
(y + 2) − y^2
]
dy Thus, the area bounded by the curves is found by the following integration:
A =
− 1
[
(y + 2) − y^2
]
dy
[
2 y +
2 y
3 y
3
] 2
− 1
=^9
( 10 pts) 2. Find the volume of the solid generated when the region between the graphs of the equations
f (x) =^1 2
over the interval [0, 2] is revolved about the x-axis. Use the disk or washer method. Include a detailed sketch with your solution. Solution: Start with a sketch.
x
y
y = x
y = 12 + x^2
(x, x)
(x, 1 /2 + x^2 )
Now, when we rotate the shaded square about the x-axis, we get a washer having an inner radius of x and an outer radius of 1/2 + x^2.
x
y
y = x
y = 12 + x^2
x 1 /2 +^ x
2
dx
In addition, the thickness is dx. Thus, the volume of the washer is the volume of the outer disk minus the volume of the inner disk:
dV = π
dx − πx^2 dx
dV = π
[(
− x^2
]
dx
x
y
(x, x)
(x, 2 x − x^2 )
1 − x dx
x − x^2
Therefore, the volume of the cylindrical shell is:
Cylindrical Shell Volume = (Circumference)(Height)(Width) = 2π(1 − x)(x − x^2 )dx
Because we fixed a value of x between 0 and 1 to start, we continue as follows:
Volume =
o
2 π(1 − x)(x 0 x^2 ) dx
= 2π
0
[
x − x^2 − x^2 + x^3
]
dx
= 2π
0
[
x^3 − 2 x^2 + x
]
dx
= 2π
[
x^4 −
x^3 +
x^2
] 1
0 = 2π
[
+^1
]
= 2π
[
+^6
]
= 2π
[
]
π 6
( 10 pts) 4. Use integration by parts using the tabular integration technique to evaluate the integral:
∫ (^1) / 2
0
sin−^1 x dx
Solution: Use tabular integration by parts.
D I
√^1
1 − x^2
x
Hence: ∫ sin−^1 x dx = x sin−^1 x −
√ x 1 − x^2
dx
= x sin−^1 x −
x(1 − x^2 )−^1 /^2 dx
= x sin−^1 x + (1 − x^2 )^1 /^2 = x sin−^1 x +
1 − x^2
Thus: ∫ (^1) / 2
0
sin−^1 x dx =
[
x sin−^1 x +
1 − x^2
] 1 / 2
0
=
2 sin
π 6
= π 12
( 10 pts) 5. Evaluate the following definite integral. (^) ∫ (^) π
0
cos^4 x dx
Solution: We’ll use the half-angle identity on this one. ∫ (^) π
0
cos^4 x dx =
∫ (^) π
0
cos^2 x
dx
∫ (^) π
0
1 + cos 2x 2
dx
∫ (^) π
0
1 + 2 cos 2x + cos^2 2 x
dx
Now the half angle identity again.
∫ (^) π
0
1 + 2 cos 2x +
1 + cos 4x 2
dx
∫ (^) π
0
(2 + 4 cos 2x + 1 + cos 4x) dx
=^1
∫ (^) π
0
(3 + 4 cos 2x + cos 4x) dx
=^1
[
3 x + 2 sin 2x +^1 4
sin 4x
]π
0
[(
3 π + 2 sin 2π +
sin 4π
]
=^3 π 8
( 10 pts) 7. Use trigonometric substitution to evaluate the following definite integral. ∫ (^1)
0
1 + x^2 dx
Solution: Let x = tan θ and dx = sec^2 θ dθ. Then: ∫ (^) √ 1 + x^2 dx =
1 + tan^2 θ (sec^2 θ dθ)
=
sec^2 θ (sec^2 θ dθ)
=
| sec θ| (sec^2 θ dθ)
Now, because the inverse tangent returns an angle in the first or fourth quadrant, we have the following picture.
√ 1 + x^2
(1, x)
θ
√ 1 + x^2
(1, x)
θ
Because the secant is positive in quadrants I and IV, we can remove the absolute value bars. Thus: ∫ (^) √ 1 + x^2 dx =
sec^3 θ dθ
This we will integrate by parts. D I
- sec θ sec^2 θ − sec θ tan θ tan θ Hence: ∫ sec^3 θ dθ = sec θ tan θ −
sec θ tan^2 θ dθ ∫ sec^3 θ dθ = sec θ tan θ −
sec θ(sec^2 θ − 1) dθ ∫ sec^3 θ dθ = sec θ tan θ −
sec^3 θ dθ +
sec θ dθ
Thus:
2
sec^3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| ∫ sec^3 θ dθ =
2 sec^ θ^ tan^ θ^ +
2 ln^ |^ sec^ θ^ + tan^ θ|
Use the picture above and note that sec θ =
1 + x^2 and tan θ = x. Hence: ∫ (^) √ 1 + x^2 dx =^1 2
sec θ tan θ +^1 2
ln | sec θ + tan θ|
=
x
1 + x^2 +
ln
1 + x^2 + x
Therefore: ∫ (^1)
0
1 + x^2 dx =
[
sec θ tan θ +
ln | sec θ + tan θ|
] 1
0 =^1 2
2 +^1
ln(1 +
( 10 pts) 8. Use trigonometric substitution to evaluate the following indefinite integral. ∫ (^2) x √ 5 − 4 x − x^2
dx
Solution: The first step is to complete the square for the denominator.
5 − 4 x − x^2 = −(x^2 + 4x − 5) = −(x^2 + 4x + 4 − 4 − 5) = −((x + 2)^2 − 9) = 9 − (x + 2)^2
Now, we can write: (^) ∫ √^2 x 5 − 4 x − x^2
dx =
√^2 x 9 − (x + 2)^2
dx
Now, we will use the following u-substitution.
u = x + 2 du = dx
Note that x = u − 2. Making these substitutions, we get: ∫ √^2 x 9 − (x + 2)^2
dx =
2( √u − 2) 9 − u^2
dx
√^2 u^ −^4 9 − u^2
dx
∫ (^2) u √ 9 − u^2
dx −
9 − u^2
dx (1)
For the first integral, let’s use these substitutions.
w = 9 − u^2 dw = − 2 u du
Now we can substitute equations (2) and (3) into equation (1) to obtain: ∫ 2 x √ 9 − (x + 2)^2
dx =
2 u √ 9 − u^2
dx −
9 − u^2
dx
= −2(9 − u^2 )^1 /^2 − 4 sin−^1
u 3
+ C
9 − u^2 − 4 sin−^1
u 3
+ C
Recall that our first substitution was u = x + 2, so our final answer is: ∫ (^2) x √ 9 − (x + 2)^2
dx = − 2
9 − u^2 − 4 sin−^1
u 3
+ C
9 − (x + 2)^2 − 4 sin−^1
x + 2 3 +^ C = − 2
5 − 4 x − x^2 − 4 sin−^1 x^ + 2 3
+ C
( 10 pts) 9. Evaluate the following improper integral. ∫ (^) ∞
e
x ln^3 x
dx
Solution: By definition: ∫ (^) ∞
e
x ln^3 x
dx = lim t→∞
∫ (^) t
e
x ln^3 x
dx
Note that if we let u = ln x and du = (1/x) dx, then: ∫ 1 x ln^3 x
dx =
u^3 du
=
u−^3 du
= − 1 2
u−^2
= − 1 2 ln^2 x Hence: ∫ (^) ∞
e
x ln^3 x
dx = lim t→∞
[
2 ln^2 x
]t
e = −
tlim→∞
[
ln^2 t
ln^2 e
]
2 tlim→∞
[
ln^2 t
]
[0 − 1]
( 10 pts) 10. Evaluate the definite integral (^) ∫ 3
0
dx (x − 1)^2 /^3
dx
Solution: The integrand is not continuous over the interval [0, 3]. It is undefined at x = 1. So we have an improper integral. The first step is to break it into two pieces.
∫ (^3)
0
dx (x − 1)^2 /^3
dx = lim b→ 1 −
∫ (^) b
0
dx (x − 1)^2 /^3
dx + lim a→ 1 +
a
dx (x − 1)^2 /^3
dx
= lim b→ 1 −
∫ (^) b
0
(x − 1)−^2 /^3 dx + lim a→ 1 +
a
(x − 1)−^2 /^3 dx
= lim b→ 1 −
[
3(x − 1)^1 /^3
]b 0
[
3(x − 1)^1 /^3
] 3
a
dx
= lim b→ 1 −
[
3(b − 1)^1 /^3 − 3(0 − 1)^1 /^3
]
[
3(3 − 1)^1 /^3 − 3(a − 1)^1 /^3
]
= lim b→ 1 −
[
3(b − 1)^1 /^3 − 3(−1)^1 /^3
]
[
3(2)^1 /^3 − 3(a − 1)^1 /^3
]
= lim b→ 1 −
[
3(b − 1)^1 /^3 − 3(−1)
]
[
3(2)^1 /^3 − 3(a − 1)^1 /^3
]
= lim b→ 1 −
[
3(b − 1)^1 /^3 + 3
]
[
3(2)^1 /^3 − 3(a − 1)^1 /^3
]
Now we can evaluate the limits. ∫ (^3)
0
dx (x − 1)^2 /^3
dx = lim b→ 1 −
[
3(b − 1)^1 /^3 + 3
]
[
3(2)^1 /^3 − 3(a − 1)^1 /^3
]
[
3(1 − 1)^1 /^3 + 3
]
[
3(2)^1 /^3 − 3(1 − 1)^1 /^3
]
= [0 + 3] +
[
3(2)^1 /^3 − 0
]
= 3 + 3(2)^1 /^3
