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Quiz #2 (v.T2): Page 1 of 4
Thursday, February 2
Short answer question
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Only answers in the boxes will
be marked. Z
(a) Evaluate (x2 + x) cos 2x3 + 3x2 dx.
Answer:
1
sin(2x3 + 3x2 ) + C
6
Solution: Substituting u = 2x3 + 3x2 , so that du = (6x2 + 6x) dx or equivalently (x2 + x) dx =
Z
2
3
(x + x) cos 2x + 3x
2
Z
dx =
cos u
du
1
1
= sin u + C = sin(2x3 + 3x2 ) + C.
6
6
6
Z
(b) Evaluate
x log x dx.
Answer:
x2
x2
log x −
+C
2
4
Solution: We integrate by parts using u = log x and dv = x dx, so that du =
Z
x log x dx =
x2
log x −
2
Z
x2 dx
x2
=
log x −
2 x
2
Z
Quiz #2 (v.T2): Page 2 of 4
Thursday, February 2
dx
x2
and v =
:
x
2
x
x2
x2
dx =
log x −
+ C.
2
2
4
du
,
6
Long answer question—you must show your work
2. 4 marks Find the area of the region above the x-axis, to the right of the curve y = 31 log x and to the left
of the curve y = log x − 2. A sketch of the region is provided below. Note that log x is the natural (base e)
logarithm of x. A calculator-ready answer is acceptable.
Answer:
2 3
3e
− e2 +
1
3
Solution: We provide two solutions. In the first we use horizontal strips, as in the figure on the left below.
In the second we use vertical strips, as in the figure on the right below.
y
y
x = e3y
y = 31 log(x)
(e3 , 1)
y = −2 + log(x)
(e3 , 1)
x = ey+2
1
x
e2
1
e2
x
Horizontal solution: The horizontal strip in the figure on the left above
runs from x = e3y to x = ey+2 . So
it has height dy and width ey+2 − e3y and hence area ey+2 − e3y dy. The area of the whole region is
Z 1
h
y+2
1
1 i1
2
e
− e3y dy = ey+2 − e3y = e3 − e2 + .
3
3
3
0
0
Marking scheme:
• 1 mark for converting the equation of one curve to x = e3y or x = ey+2 or x = e2 ey . (This can be
implicit in the integrand.)
• 1 mark for the correct integrand.
• 1 mark for the correct limits of integration.
• 1 mark for the correct answer to their integral.
Vertical solution: The equations of the curves at the top of our region and at the bottom of our region are
(
0,
if 1 ≤ x ≤ e2 ,
1
y = B(x) =
y = T (x) = log x,
3
log x − 2, if e2 ≤ x ≤ e3 .
So the area is
Z e3
Z
T (x) − B(x) dx =
e2
Z
e2
1
T (x) − B(x) dx +
1
=
h1
Z
i
log x dx +
e3
Z
h1
e3
e2
T (x) − B(x) dx
i
log x − log x − 2 dx
3
3
e2
1
e3
ie2 2 h
i e3
1h
= x log x − x
− x log x − x 2 + 2x 2
3
3
1
e
e
1 2
2
1
1 2
2
2
3
3
2
=
2e − e + −
3e − e +
2e − e2 + 2e3 − 2e2 = + e3 − e2 .
3
3 3
3
3 3
R
That log x dx = x log x − x + C can be found by integration by parts (or by guessing).
Marking scheme:
•
•
•
•
1
1
1
1
mark for splitting up the domain of integration
mark for the limits
marks for the integrands
mark for the answer.
Quiz #2 (v.T2): Page 3 of 4
Thursday, February 2
Long answer question—you must show your work
3. 4 marks Let R be the region above the x-axis, to the right of the curve y 2 = x − 1, and to the left of the
curve y 2 = 3 − x. Using disks and/or washers, find the volume of the solid obtained by rotating the region R
about the line x = 1. A calculator-ready answer is acceptable.
Answer:
8
3π
Solution: The curves are x = y 2 + 1, which is a rightward opening parabola, and x = 3 − y 2 , which is a
leftward opening parabola. They meet when
y 2 + 1 = 3 − y 2 ⇐⇒ 2y 2 = 2 ⇐⇒ y 2 = 1 ⇐⇒ y = ±1
Here is a sketch of the region.
(2, 1)
y
x = 3 − y2
1
2
3 x
x=y +1
x=1
When it is rotated about the line x = 1, the horizontal strip in the figure sweeps out a washer with thickness
dy, inner radius rin = y 2 + 1 − 1 = y 2 and outer radius rout = 3 − y 2 − 1 = 2 − y 2 and hence of volume
2
2
2
π rout
− rin
dy = π (2 − y 2 ) − (y 2 )2 dy = π 4 − 4y 2 dy
So the volume of the solid is
Z 1
Z
2
π 4 − 4y dy = 4π
0
0
1
1
1
8
y3
2
= 4π 1 −
= π
1 − y dy = 4π y −
3 0
3
3
Marking scheme:
• 2 marks for the correct washer volume
• 1 mark for the correct limits of integration
• 1 mark for the correct answer to their integral
Quiz #2 (v.T2): Page 4 of 4
Thursday, February 2