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Quiz #3 (v.M3): 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
sec x
Answer:
1
1
+ cos x + C = sec x +
+C
cos x
sec x
Solution: Substituting u = cos x, so that du = − sin x dx and sin2 x = 1 − u2
Z
Z
Z
Z
1
tan3 x
sin3 x
sin2 x
1 − u2
du = + u + C
dx =
dx
=
sin
xdx
=
−
3
2
2
sec x
cos x/ cos x
cos x
u
u
1
=
+ cos x + C
cos x
Alternatively, we can also substitute u = sec x, du = sec x tan x dx, tan2 x = sec2 x − 1 = u2 − 1,
Z
Z 2
Z
tan2 x
u −1
1
tan3 x
dx =
sec
x
tan
x
dx
=
du = u + + C
sec x
sec2 x
u2
u
1
+C
= sec x +
sec x
√
2 2
Z
√
(b) Evaluate
0
1
dx. Simplify your answer fully.
16 − x2
Answer:
π
4
√
Solution: Making the substitution x = 4 sin u, dx = 4 cos u du, 16 − x2 = 4 cos u
Z
0
√
2 2
1
√
dx =
16 − x2
Z
√
arcsin( 2/2)
arcsin(0)
√
1
π
4 cos u du = arcsin( 2/2) − arcsin(0) = .
4 cos u
4
Quiz #3 (v.M3): Page 2 of 4
Friday, February 17
Long answer question—you must show your work
Z
√
2 3
2. 4 marks Calculate
2
4x + 4
dx.
x3 + 4x
Solution: We decompose the integrand using partial fractions, writing
4x + 4
4x + 4
A Bx + C
=
= + 2
.
3
2
x + 4x
x(x + 4)
x
x +4
Multiplying through by x(x2 + 4), we get
4x + 4 = A(x2 + 4) + (Bx + C)x = (A + B)x2 + Cx + 4A.
One way to determine A, B and C is to equate coefficients, getting the equations
0 = A+B
4 = C
4 = 4A
(coefficients of x2 ),
(coefficients of x),
(constant terms).
Solving this system of equations, we get A = 1, B = −1 and C = 4. Thus
Z
2
√
2 3
4x + 4
dx =
x3 + 4x
Z
2
Z
=
2
√
2 3
√
2 3
1
x
4
−
+
x x2 + 4 x2 + 4
dx
x
1
1
− 2
+ x 2
x x + 4 (( 2 ) + 1)
dx
2√3
2
x
1
= log |x| − log x + 4 + 2 arctan
2
2
2
√
1
2π
1
π
= log(2 3) − log(16) +
− log(2) + log(8) −
2
3
2
2
π 1
1
= − log(2) + log(3).
6
2
2
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.M3): Page 3 of 4
Friday, February 17
Long answer question—you must show your work
Z
0
3. (a) 1 mark Estimate
x
−ee dx using the Trapezoid Rule and n = 3 subintervals.
−1
Solution: We have ∆x =
T3 =
1
3
and so
1 e−1
−2/3
−1/3
1
−e
− 2ee
− 2ee
−e .
f (−1) + 2f − 32 + 2f − 31 + f (0) =
6
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
x
x
x
d3
d2
(−ee ) = −ee +x (ex + 1) and
(−ee ) = −ee +x 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 negative and
decreasing on [−1, 0] and takes on its maximum absolute value at x = 0. Thus on [−1, 0] we have
|f 00 (x)| ≤ 2e. By the formula given, the total error for the estimate in part (a) is bounded by
2e
2e(0 − (−1))3
=
.
2
12(3 )
108
Marking scheme:
• 1 mark for coming up with a bound on the “M ” term in the formula that uses the facts that
d2
ex
dx2 (−e ) is negative and decreasing (both facts must be used)
• 1 mark for coming up with a correct bound on the estimate in part (a) using whatever value
of M was given above
Z
0
(c) 1 mark Is the estimate in part (a) greater than, equal to, or less than the actual value of
x
−ee dx?
−1
Justify your answer in one to three sentences.
x
2
x
d
e
)=
Solution: The estimate is less than the actual value of the integral. Since −ee < 0 and dx
2 (−e
ex +x
x
ex
−e
(e + 1) < 0 for all x, y = −e is concave down on [0, 1] (and indeed everywhere), and the
trapezoids used in the estimate lie below the curve.
Marking scheme: 1 mark for concluding that the estimate is an underestimate by observing that
x
−ee is concave down.
Quiz #3 (v.M3): Page 4 of 4
Friday, February 17