First Name:
Last Name:
Student-No:
Section:
Grade:
This page will be overwritten with the fancy auto-multiple-choice front page.
Quiz #3 (v.M1): 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
sec4 x
Answer: −
cos2 x cos4 x
1
1
+
+C =−
+
+C
2
4
2 sec2 x 4 sec4 x
Solution: Substituting u = cos x, so that du = − sin x dx and sin2 x = 1 − u2
Z
Z
Z
Z
tan3 x
u2
u4
sin3 x
2
2
dx
=
dx
=
sin
x
cos
x
sin
xdx
=
−
(1
−
u
)u
du
=
−
+
+C
sec4 x
cos3 x/ cos4 x
2
4
cos2 x cos4 x
=−
+
+C
2
4
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
1
tan3 x
tan2 x
u −1
dx
=
sec
x
tan
x
dx
=
du = − 2 + 4 + C
4
5
5
sec x
sec x
u
2u
4u
1
1
=−
+
+C
2 sec2 x 4 sec4 x
√
2 3
Z
√
(b) Evaluate
0
1
dx. Simplify your answer fully.
16 − x2
Answer:
π
3
√
Solution: Making the substitution x = 4 sin u, dx = 4 cos u du, 16 − x2 = 4 cos u
Z
0
√
2 3
1
√
dx =
16 − x2
Z
√
arcsin( 3/2)
arcsin(0)
√
1
π
4 cos u du = arcsin( 3/2) − arcsin(0) = .
4 cos u
3
Quiz #3 (v.M1): Page 2 of 4
Friday, February 17
Long answer question—you must show your work
√
Z
3
2. 4 marks Calculate
1
x+1
dx.
x3 + x
Solution: We decompose the integrand using partial fractions, writing
x+1
x+1
A Bx + C
=
= + 2
.
3
2
x +x
x(x + 1)
x
x +1
Multiplying through by x(x2 + 1), we get
x + 1 = 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
1
= A+B
= C
= A
(coefficients of x2 ),
(coefficients of x),
(constant terms).
Solving this system of equations, we get A = 1, B = −1 and C = 1. Thus
√
Z
1
3
√
x+1
dx
x3 + x
3
1
x
1
− 2
+ 2
dx
x x +1 x +1
1
√3
2
1
=
log |x| − log x + 1 + arctan(x) 2
1
√ 1
π
1
π
3 − log(4) + − 0 + log(2) − .
= log
2
3
2
4
Z
=
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.M1): Page 3 of 4
Friday, February 17
Long answer question—you must show your work
Z
3. (a) 1 mark Estimate
1
ee
−x
dx using the Trapezoid Rule and n = 3 subintervals.
0
Solution: We have ∆x =
T3 =
1
3
and so
1
f (0) + 2f
6
1
3
+ 2f
2
3
1
−1/3
−2/3
−1
e + 2ee
+ 2ee
+ ee
.
+ f (1) =
6
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, 1] and takes on its maximum value at x = 0. On [0, 1], we have |f 00 (x)| ≤ 2e. By the formula
given, the total error for the estimate in part (a) is bounded by
2e
2e(1 − 0)3
=
.
12(32 )
108
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
1
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, 1] (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.M1): Page 4 of 4
Friday, February 17