Midterm Exam #1—Math 101, Section 207
January 30, 2015
Duration: 50 minutes
Name:
Student Number:
Do not open this test until instructed to do so! This exam should have 8 pages, including this cover sheet.
No textbooks, notes, calculators, or other aids are allowed; phones, pencil cases, and other extraneous items
cannot be on your desk. Turn off cell phones and anything that could make noise during the exam.
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be given
for the answer without the correct accompanying work. Problems 5–7 are long-answer: give complete
arguments and explanations for all your calculations—answers without justifications will not be marked.
Continue on the back of the page if you run out of space.
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Problem Out of Score Problem Out of Score
1
6
5
8
2
6
6
8
3
6
7
8
4
3
Total
45
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 2 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
1a. [3 pts] Set up (but do not evaluate) a right Riemann sum approximating, with N rectangles,
the area under the graph of f (x) = 1 + 6x between x = 0 and x = 2.
Solution: We take a right Riemann sum with N subintervals, so that ∆xi =
2i
xi = x∗i = N
. With this we may write
N
X
i=1
f (x∗i )∆xi
=
N
X
i=1
2i
)( N2 )
f(N
=
N
X
(1 +
2
N
and
12i
)( N2 ).
N
i=1
Grading scheme:
• 1pt. for the correct form of the Riemann sum.
• 1pt. for the correct interval length N2 .
• 1pt. for choosing x∗i correctly.
1b. [3 pts] Find the exact value of the area under the graph of f (x) = 1 + 6x between 0 and 2.
Z
2
1 + 6x dx. An anti-derivative
Solution: We seek the value of the definite integral
0
of f (x) is F (x) = x + 3x2 , so by the fundamental theorem of calculus we have
Z 2
(1 + 6x) dx = [F (x)]2x=0 = F (2) − F (0) = (2 + 3(2)2 ) − (0 + 3(0)2 ) = 14.
0
Grading scheme:
• 1pt. for indicating how to find the area (by integration or limit of Riemann sums).
• 1pt. for the correct anti-derivative or limit.
• 1pt. for the correct value.
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 3 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
Z cos x
d
dt
2a. [3 pts] Determine
.
dx 0
1 + t2
Solution: We use the
rule, together with the first fundamental theorem of
R x chain
dt
calculus. Let f (x) = 0 1+t
and
let g(x) = cos x. Then clearly g 0 (x) = − sin x and
2
1
by FTC I we know that f 0 (x) = 1+x
2 . Then by the chain rule
Z cos x
1
− sin x
d
dt
d
0
0
f (g(x)) = f (g(x))g (x) =
(− sin x) =
.
=
2
2
dx 0
1+t
dx
1 + g(x)
1 + cos2 x
Grading scheme:
• 1pt. for indicating FTC 1.
• 1pt. for indicating the chain rule.
• 1pt. for the correct answer.
Z
2b. [3 pts] Compute the definite integral
1
eπ/2
sin(ln x)
dx.
x
Solution: We use the substitution u = ln x so that du = dx
. Then since u = 0 when
x
π
π/2
x = 1 and u = 2 when x = e , we have
Z π/2
Z π/2
Z eπ/2
sin u
sin(ln x)
sin u du
dx =
(x du) =
x
x
0
0
1
π/2
= [− cos u]u=0 = − cos( π2 ) − (− cos 0) = 1.
Grading scheme:
• 1pt. for the substitution u = ln x.
• 1pt. for correctly dealing with the limits of integration and integrand.
• 1pt. for the correct answer.
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 4 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
3a. [3 pts] Find the area of the region enclosed by the curves y = x2 and y = 4x − x2 .
Solution: We note that the curves intersect when x2 = 4x − x2 , which is exactly
when 2x2 = 4x. One solution is x = 0, and for x 6= 0 we divide by x to obtain 2x = 4
or x = 2. On the interval [0, 2], the curve y = 4x − x2 lies above the other curve (we
can check this by, say, plugging in x = 1), so the area we want is given by the integral
Z 2
Z 2
2
2
4x − 2x2 dx = [2x2 − 32 x3 ]2x=0
(4x − x ) − x dx =
0
0
8
= (2(2)2 − 23 (2)3 ) − (2(0)2 − 23 (0)3 ) = .
3
Grading scheme:
• 1pt. for finding the points of intersection.
• 1pt. for indicating which curve is on top, and why.
• 1pt. for the correct calculation.
3b. [3 pts] On T WIN E ARTH where acceleration due to gravity is exactly 10 m/s2 , a 4m long,
20kg steel cable of uniform density is lying on the ground. Determine the work done by
lifting one end of the cable to a height of 4m (so that the cable is fully vertical).
Solution: The linear density of the cable is 204 mkg = 5 kg/m, and so the mass of x
metres of cable is 5x kg. The force exerted by gravity on the vertical length of cable x
metres long is exactly (mass)(g) = (5x kg)(10 m/s2 ) = 50x Newtons. Thus the work
done is exactly
Z 4
50x dx = [25x2 ]4x=0 = 25(4)2 − 25(0)2 = 400 Joules.
W =
0
Grading scheme:
• 1pt. for finding linear density.
• 1pt. for finding the force function.
• 1pt. for the correct calculation.
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 5 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
4. [3 pts] √
Find the volume of the solid obtained by rotating about the x-axis the region under the
curve y = x − 1 from x = 1 to 6.
Solution: The area of a circle of radius r is πr2 , and so area ofZa cross-section at point x is
6
√
exactly π( x − 1)2 = π(x−1). Thus the volume that we seek is
π(x−1) dx. We integrate
the polynomial:
Z 6
Z
V =
π(x − 1) dx =
1
1
6
Z
πx dx −
1
Grading scheme:
• 1pt. for the correct area function.
• 1pt. for the correct limits of integration.
• 1pt. for the correct calculation.
1
6
2
π dx = [ πx2 ]6x=1 − [πx]6x=1 =
25π
.
2
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 6 of 8
Problems 5–7 are long-answer: give complete arguments and explanations for all your calculations—
answers without justifications will not be marked.
5.
Z
51
sin(sin x)dx.
(a) [3 pts] Without computing any Riemann sums or integrals, find the value of
−51
Solution: The function sin x is odd, and therefore so is the function sin(sin x):
sin(sin(−x)) = sin(− sin x) = − sin(sin x).
Moreover sin(sin x) is continuous
everywhere; as seen in lecture, for any continuous,
Z
a
f (x) dx = 0 for any a ∈ R, and so in particular we
odd function f (x) one has
−a
Z 51
know that
sin(sin x)dx = 0.
−51
Grading scheme:
• 1pt. for citing the result about integrals of odd functions on [−a, a].
• 1pt. for stating that sin(sin x) is odd (and contnuous).
• 1pt. for justifying why sin(sin
is odd.
Z x)
51
sin(sin x)
(b) [2 pts] Explain why the integral
dx cannot be computed using the method
x2
−51
of part (a).
x)
Solution: Since the function sin(sin
is not continuous on the interval [−51, 51] (it
x2
is discontinuous at x = 0), the result of the theorem cannot be applied.
Grading scheme:
• 1pt. for noting that the function is not continuous on the interval.
• 1pt. for stating that the theorem cannot be
Z applied (as it assumesZcontinuity).
16
(c) [3 pts] Suppose f (x) is continuous and that
2
2x3 f (x4 )dx.
f (x)dx = 1. Find
0
0
Solution: We use the substitution u = x4 so that du = 4x3 dx. Then u = 16 when
x = 2 and u = 0 when x = 0, so
Z 2
Z 16
Z
du
1
1 16
1
3
4
3
2x f (u) 3 =
2x f (x )dx =
f (u)du = (1) = .
4x
2 0
2
2
0
0
Grading scheme:
• 1pt. for the substitution u = x4 .
• 2pts. for the correct calculation
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 7 of 8
6.
(a) [2 pts] Find the points of intersection of the two curves y1 = 12 x2 and y2 = 2x − x2 .
Solution: We equate the two expressions and solve the resulting quadratic equation:
3
1 2
x = 2x − x2 ⇔ x2 − 2x = 0 ⇔ x(3x − 4) = 0.
2
2
Clearly the only solutions are x = 0 and x = 34 .
Grading scheme:
• 1pt. for equating the functions to find their intersection points.
• 1pt. for the values 0 and 43 .
(b) [6 pts] Find the volume of the solid obtained by rotating about the x-axis the region
bounded by the two curves in part (a); your answer does not need to be simplified. (Hint:
Recall the binomial theorem (a + b)2 = a2 + 2ab + b2 )
Solution: By part (a) we know that the two curves intersect at 0 and 43 , and by
checking the value x = 1 we see that the curve y2 lies above y1 (since y1 (1) = 21 <
1 = y2 (1)). The cross-sectional area is therefore
1
3
A(x) = πR2 − πr2 = πy22 − πy12 = π((4x2 − 4x3 + x4 ) − x4 ) = π( x4 − 4x3 + 4x2 ).
4
4
Thus
Z
Z
4/3
4/3
V =
A(x)dx =
π( 34 x4 − 4x3 + 4x2 )dx
0
0
4/3
3 5
4 3
4 4 3
3 4 5
4
4 4
=π
x −x + x
( ) − (3) + (3) .
=π
20
3
20 3
3
x=0
Grading scheme:
• 1pt. for finding which function is on top.
• 1pt. for using the correct area function.
• 1pt. for the correct limits of integration.
• 1pt. for the correct volume integral.
• 1pt. for a valid anti-derivative.
• 1pt. for the correct answer.
Midterm Exam #1—Math 101, Section 207—January 30, 2015
page 8 of 8
7.
π/2
Z
Z
0
0
(Hint: recall the identity sin( π2 − x) = cos x).
π/2
f (cos x) dx.
f (sin x) dx =
(a) [4 pts] Suppose f (x) is continuous. Prove that
Solution: Using the hint, we substitute u = π2 − x so that du = −dx and we have
Z π/2
Z π/2
Z 0
π
f (cos x) dx =
f (sin( 2 − x)) dx =
f (sin u)(−du)
0
0
Z
0
=−
Z
π/2
f (sin u) du.
f (sin u) du =
0
π/2
π/2
π/2
Z
f (sin u) du =
We obviously have
follows.
π/2
Z
f (sin x) dx, from which the equality
0
0
Grading scheme:
• 1pt. for substituting u = π2 − x.
• 2pts. for the correct calculation.
• 1pt. for rigor/clarity.
π/2
Z
2
Z
sin x dx and
(b) [4 pts] Use the result of part (a) to find
0
π/2
cos2 x dx. You may
0
assume the result of part (a) even if you did not prove it. (Hint: consider what f (x) should
be in order to apply part (a))
Solution: By the identity sin2 x + cos2 x = 1 we have
Z π/2
Z π/2
Z π/2
Z π/2
π
2
2
2
=
du =
sin u + cos u du =
sin u du +
cos2 u du.
2
0
0
0
0
With f (x) = x2 we can write
Z π/2
Z
f (sin u) du +
π/2
π
,
2
0
0
and since both integrals on the left are equal by part (a) we know that
Z π/2
Z π/2
1 π π
2
2
cos u du =
sin u du =
= .
2 2
4
0
0
f (cos u) du =
Grading scheme:
• 1pt. for stating sin2 x + cos2 x = 1.
• 1pt. for applying the result from part (a)/stating what f (x) should be.
• 1pt. for calcultions.
• 1pt. for finding the correct value.