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Quiz #3 (v.M2): Page 1 of 4
Friday, February 17
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
tan3 x
dx.
(a) Evaluate
sec2 x
Answer: − log | cos x| +
cos2 x
1
+ C = log | sec x| +
+C
2
2 sec2 x
Solution: Substituting u = cos x, so that du = − sin x dx and sin2 x = 1 − u2
Z
Z
Z
Z
tan3 x
u2
sin3 x
sin2 x
1 − u2
dx
=
dx
=
sin
xdx
=
−
du
=
−
log
|u|
+
+C
sec2 x
cos3 x/ cos2 x
cos x
u
2
cos2 x
= − log | cos x| +
+C
2
Alternatively, we can also substitute u = sec x, du = sec x tan x dx, tan2 x = sec2 x − 1 = u2 − 1,
Z
Z
Z 2
1
tan3 x
tan2 x
u −1
dx
=
sec
x
tan
x
dx
=
du = log |u| + 2 + C
2
3
3
sec x
sec x
u
2u
1
= log | sec x| +
+C
2 sec2 x
Z
(b) Evaluate
2
√
0
1
dx. Simplify your answer fully.
16 − x2
Answer:
π
6
√
Solution: Making the substitution x = 4 sin u, dx = 4 cos u du, 16 − x2 = 4 cos u
Z
0
2
√
1
dx =
16 − x2
Z
arcsin(1/2)
arcsin(0)
1
π
4 cos u du = arcsin(1/2) − arcsin(0) = .
4 cos u
6
Quiz #3 (v.M2): Page 2 of 4
Friday, February 17
Long answer question—you must show your work
Z
2. 4 marks Calculate
1
√
1/ 3
x−2
dx.
x3 + x
Solution: We decompose the integrand using partial fractions, writing
x−2
x−2
A Bx + C
=
= + 2
.
3
2
x +x
x(x + 1)
x
x +1
Multiplying through by x(x2 + 1), we get
x − 2 = A(x2 + 1) + (Bx + C)x = (A + B)x2 + Cx + A.
One way to determine A, B and C is to equate coefficients, getting the equations
0 =
1 =
−2 =
A+B
C
A
(coefficients of x2 ),
(coefficients of x),
(constant terms).
Solving this system of equations, we get A = −2, B = 2 and C = 1. Thus
Z 1 Z 1
x−2
−2
2x
1
dx =
+ 2
+
dx
√
√
3
x
x + 1 x2 + 1
1/ 3
1/ 3 x + x
2
1
=
−2 log |x| + log x + 1 + arctan(x) 1/√3
π
1
4
π
+
.
= log(2) + − −2 log √
+ log
4
3
6
3
Marking scheme:
• 1 mark for the correct partial fractions decomposition formula (without solving for the constants)
• 1 mark for solving the above for the correct constants
• 1 mark for antidifferentiating the decomposed integrand, even if the decomposition is incorrect, but
not if the decomposition consists of only one or two terms
• 1 mark for a final answer consistent with earlier work
Quiz #3 (v.M2): Page 3 of 4
Friday, February 17
Long answer question—you must show your work
Z
3. (a) 1 mark Estimate
2
ee
−x
dx using the Trapezoid Rule and n = 4 subintervals.
0
Solution: We have ∆x =
T4 =
1
f (0) + 2f
4
1
2
1
2
and so
+ 2f (1) + 2f
3
2
1
−1/2
−1
−3/2
−2
e + 2ee
+ 2ee + 2ee
+ ee
.
+ f (2) =
4
Marking scheme: 1 mark for the correct answer
Z
(b) 2 marks Suppose an integral
b
f (x) dx is estimated using the Trapezoid Rule and n subintervals. If
a
3
(b−a)
|f 00 (x)| ≤ M for a ≤ x ≤ b, then the total error is bounded by M12n
. Use this fact to find a bound on
2
the total error for the estimate in part (a). You may also use without proof the facts that
−x
d3 e−x
d2 e−x
e−x −2x
x
e
=
e
(e
+
1)
and
e
= −ee −3x e2x + 3ex + 1 .
2
3
dx
dx
You must justify your choice of M .
−x
Solution: Let f (x) = ee . Since f 00 (x) > 0 and f (3) (x) < 0 for all x, f 00 (x) is positive and decreasing
on [0, 2] and takes on its maximum value at x = 0. On [0, 2], we have |f 00 (x)| ≤ 2e. By the formula
given, the total error for the estimate in part (a) is bounded by
16e
2e(2 − 0)3
=
.
12(42 )
192
Marking scheme:
• 1 mark for coming up with a bound on the “M ” term in the formula that uses the facts that
d2 e−x
is positive and decreasing (both facts must be used)
dx2 e
• 1 mark for coming up with a correct bound on the estimate in part (a) using whatever value
of M was given above
2
Z
−x
ee
(c) 1 mark Is the estimate in part (a) greater than, equal to, or less than the actual value of
dx?
0
Justify your answer in one to three sentences.
−x
2
−x
d
e
Solution: The estimate is greater than the actual value of the integral. Since ee > 0 and dx
=
2e
−x
−x
e −2x
x
e
e
(e + 1) > 0 for all x, y = e
is concave up on [0, 2] (and indeed everywhere), and the
trapezoids used in the estimate lie above the curve.
Marking scheme: 1 mark for concluding that the estimate is an overestimate by observing that
−x
ee is concave up.
Quiz #3 (v.M2): Page 4 of 4
Friday, February 17