Download Test 1

MAT 266 Test 1 SOLUTIONS, FORM A
1. [20 points] Perform the appropriate trigonometric substitution on the following integral. Do NOT try
to find the resulting integral.
Z √ 2
x − 9 + 5x
√
dx
x
e − x2 − 9
p
√
Solution: The presence ofpthe x2 − 9 formula suggests that you let x = 9 · sec θ = 3 sec θ. Then
dx = 3 sec θ tan θ dθ and x2 − 9 = 3 tan θ. Substituting these into the formula produces:
Z
3 tan θ + 5 · 3 sec θ
· 3 sec θ tan θ dθ
e3 sec θ − 3 tan θ
Grading (roughly): +9 points for the substitutions for x, dx, and
tion. Grading for common mistakes: −5 points for not replacing dx.
√
·; +11 points for substitu-
2. [15 points] Find the FORM of the Partial Fraction Decomposition of the rational function below. You
do not need to determine the coefficients.
(x + 4)(x − 2)2 (x2 + 5x + 8)
x3 (x + 3)(x2 − 5x + 8)2
B
C
D
Ex + F
Gx + H
A
+ 2+ 3+
+ 2
+ 2
. Note that x2 − 5x + 8 is a quadx
x
x
x + 3 x − 5x + 8 (x − 5x + 8)2
ratic polynomial with no real roots.
Grading: Done on a 0−3−5−7−10−12−15-point basis (these were the grades usually given);
generally, you went from 15 to 12 if you made one mistake, 12 to 10 if you made another, etc.
Solution:
2
MAT 266 Test 1 SOLUTIONS, FORM A
3. Find the following integrals.
Z
a. [15 points]
x2 e5x dx
Solution: This is done using Integration by Parts twice:
Z
2 5x
x e
e5x
dx = x ·
−
5
2
=
=
x2 e5x
2
−
5
5
Z
e5x
x2 e5x
2
2x ·
dx =
−
5
5
5
x2 e5x
2
−
5
5
x·
e5x
−
5
Z
xe5x
e5x
−
5
25
1·
xe
e5x
dx
5
#
v 0 = e5x
e5x
u0 = 2x v =
5
#
"
u = x v 0 = e5x
e5x
u0 = 1 v =
5
"
Z
5x
dx
u = x2
+C
Grading for common mistakes: +5 points (total) for starting with u = e5x and v 0 = x2 ;
+7 points (total) for only one integration-by-parts.
Z
1
b. [15 points]
dx
x2 − 4x
Solution: This is a rational function, so do it with partial fractions:
1
1
A
B
=
= +
x2 − 4x
x(x − 4)
x
x−4
1 = A(x − 4) + Bx
1 = −4 · A
1=4·B
⇒
⇒
A=−
B=
1
4
1
4
(x = 0)
(x = 4)
Then
Z
1
dx =
x2 − 4x
Z
1 1 1
1
1
1
− · + ·
dx = − ln x + ln(x − 4) + C.
4 x 4 x−4
4
4
Grading for common mistakes: +5 points (total) for ln(x2 −4x); −3 points for factoring x2 −4x
incorrectly.
Z
c. [15 points]
sin5 x cos2 x dx
Solution: This integral has trig functions, so it is a trig integral. This one can be done by setting
up the substitution u = cos x (since sin x has an odd power, and cos x has an even power); the rest
of the substitution is du = − sin x dx, and sin2 x = 1 − cos2 x.
Z
Z
Z
sin5 x cos2 x dx = sin4 x cos2 x · sin x dx = (1 − cos2 x)2 cos2 x · sin x dx
Z
Z
2 2
2
= (1 − u ) · u · (−1) du = −u2 + 2u4 − u6 du
=
−u3
2u5
u7
− cos3 x 2 cos5 x cos7 x
+
−
+C =
+
−
+ C.
3
5
7
3
5
7
Grading for common mistakes: +7 points (total) for using power-reducing formulas.
3
MAT 266 Test 1 SOLUTIONS, FORM A
Z
5
ln(2 + sin x) dx cannot be determined. Approxi-
4. [20 points] The exact value of the definite integral
2
mate this integral using the Trapezoid Rule and Simpson’s Rule for n = 4. (You will only need to write
out the formulas; make sure you identify which is which.)
Solution: Before providing these formulas, a few numbers need to be calculated:
∆x =
b−a
5−2
=
= 0.75,
n
4
and x0 , x1 , x2 , x3 , x4 is the sequence 2, 2.75, 3.5, 4.25, 5.
The Trapezoid Rule states that
Z
5
ln(2 + sin x) dx ≈
2
=
∆x
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + f (x4 )]
2
0.75
· [ln(2 + sin 2) + 2 ln(2 + sin 2.75) + 2 ln(2 + sin 3.5) + 2 ln(2 + sin 4.25) + ln(2 + sin 5)]
2
≈ 1.516527504.
Simpson’s Rule states that
Z
5
ln(2 + sin x) dx ≈
2
=
∆x
[f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 )]
3
0.75
· [ln(2 + sin 2) + 4 ln(2 + sin 2.75) + 2 ln(2 + sin 3.5) + 4 ln(2 + sin 4.25) + ln(2 + sin 5)]
3
≈ 1.496386232.
Grading: +2 points for ∆x, +4 points for x0 , x1 , x2 , x3 , x4 , +5 points for each basic formula,
+2 points for each substitution. Grading for common mistakes: −1 points for a calculation done
in degree mode; −3 points for keeping f in the final answer.
4
MAT 266 Test 1 SOLUTIONS, FORM B
1. [20 points] Perform the appropriate trigonometric substitution on the following integral. Do NOT try
to find the resulting integral.
p
Z sin
4 − x2 + 5
√
dx
1 − x 4 − x2
p
√
2
Solution: The presence
p of the 4 − x formula suggests that you let x = 4 · sin θ = 2 sin θ. Then
dx = 2 cos θ dθ and 4 − x2 = 2 cos θ. Substituting these into the formula produces:
Z
sin (2 cos θ) + 5
· 2 cos θ dθ
1 − 2 sin θ · 2 cos θ
Grading (roughly): +9 points for the substitutions for x, dx, and
tion. Grading for common mistakes: −5 points for not replacing dx.
√
·; +11 points for substitu-
2. [15 points] Find the FORM of the Partial Fraction Decomposition of the rational function below. You
do not need to determine the coefficients.
(x + 5)2 x(x2 − 3x + 9)2
(x + 1)3 (x − 5)(x2 + 3x + 9)2
Solution:
A
B
Ex + F
Gx + H
C
D
+
+
+
+
+
. (Note that
x + 1 (x + 1)2
(x + 1)3
x − 5 x2 + 3x + 9 (x2 + 3x + 9)2
x2 + 3x + 9 is a quadratic polynomial with no real roots.)
Grading: Done on a 0−3−5−7−10−12−15-point basis (these were the grades usually given);
generally, you went from 15 to 12 if you made one mistake, 12 to 10 if you made another, etc.
2
MAT 266 Test 1 SOLUTIONS, FORM B
3. Find the following integrals.
Z
1
a. [15 points]
dx
x2 + 3x
Solution: This is a rational function, so do it with partial fractions:
x2
1
1
A
B
=
= +
+ 3x
x(x + 3)
x
x+3
1 = A(x + 3) + Bx
1=3·A
⇒
1 = −3 · B
A=
⇒
1
3
(x = 0)
B=−
1
3
(x = −3)
Then
Z
1
dx =
2
x + 3x
Z
1 1 1
1
1
1
· − ·
dx =
ln x − ln(x + 3) + C.
3 x 3 x+3
3
3
Grading for common mistakes: +5 points (total) for ln(x2 + 3x).
Z
b. [15 points]
tan3 x sec80 x dx
Solution: This integral has trig functions, so it is a trig integral. This one can be done either by
setting up the substitution u = tan x or u = sec x; after the substitution u = sec x, the new integral
is easier to do. Note that, since u = sec x, du = sec x tan x dx, and tan2 x = sec2 x − 1.
Z
Z
Z
tan3 x sec80 x dx = tan2 x sec79 x · sec x tan x dx = (sec2 x − 1) sec79 x · sec x tan x dx
Z
=
Z
c. [15 points]
(u2 − 1)u79 du =
Z
u81 − u79 du =
u82
u80
sec82 x sec80 x
−
+C =
−
+C
82
80
82
80
x2 e−4x dx
Solution: This is done using Integration by Parts twice:
Z
2 −4x
x e
e−4x
−
dx = x ·
−4
2
=
=
e−4x
−x2 e−4x
1
2x ·
dx =
+
−4
4
2
Z
−x2 e−4x
1
+
4
2
1
−x2 e−4x
+
4
2
x·
e−4x
−
−4
Z
1·
−xe−4x
e−4x
−
4
16
e−4x
dx
−4
#
v 0 = e−4x
−4x
e
u0 = 2x v =
−4
#
"
u = x v 0 = e−4x
e−4x
u0 = 1 v =
−4
"
Z
xe
−4x
dx
u = x2
+C
Grading for common mistakes: +5 points (total) for letting u = e−4x and v 0 = x2 .
3
MAT 266 Test 1 SOLUTIONS, FORM B
Z
5
tan(5 + ln x) dx cannot be determined. Approxi-
4. [20 points] The exact value of the definite integral
3
mate this integral using the Trapezoid Rule and Simpson’s Rule for n = 4. (You will only need to write
out the formulas; make sure you identify which is which.)
Solution: Before providing these formulas, a few numbers need to be calculated:
∆x =
b−a
5−3
=
= 0.5,
n
4
and x0 , x1 , x2 , x3 , x4 is the sequence 3, 3.5, 4, 4.5, 5.
The Trapezoid Rule states that
Z
5
tan(5 + ln x) dx ≈
3
=
∆x
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + f (x4 )]
2
0.5
· [tan(5 + ln 3) + 2 tan(5 + ln 3.5) + 2 tan(5 + ln 4) + 2 tan(5 + ln 4.5) + tan(5 + ln 5)]
2
≈ 0.1867116031.
Simpson’s Rule states that
Z
5
tan(5 + ln x) dx ≈
3
=
∆x
[f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 )]
3
0.5
· [tan(5 + ln 3) + 4 tan(5 + ln 3.5) + 2 tan(5 + ln 4) + 4 tan(5 + ln 4.5) + tan(5 + ln 5)]
3
≈ 0.1892353202.
Grading: +2 points for ∆x, +4 points for x0 , x1 , x2 , x3 , x4 , +5 points for each basic formula,
+2 points for each substitution. Grading for common mistakes: −1 points for a calculation done
in degree mode; −3 points for keeping f in the final answer.
4