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Math 106: Review for Exam II - SOLUTIONS
INTEGRATION TIPS
- Substitution: usually let w = an inside function, especially if w′^ is also present in the integrand
- Parts:
u dv = uv −
v du or
uv′^ dx = uv −
u′v dx
How to choose which part is u? Let u be the part that is higher up in the LIATE mnemonic below. (The mnemonics ILATE and LIPET will work equally well if you have learned one of those instead; in the latter A is replaced by P, which stands for “polynomial”.) Logarithms (such as ln x) Inverse trig (such as arctan x, arcsin x) Algebraic (such as x, x^2 , x^3 + 4) Trig (such as sin x, cos 2x) Exponentials (such as ex^ , e^3 x)
- Rational Functions (one polynomial divided by another): if the degree of the numerator is greater than or equal to the degree of the numerator, do long division then integrate the result. Partial Fractions: here’s an illustrative example of the setup.
3 x^2 + 11 (x + 1)(x − 3)^2 (x^2 + 5)
A
x + 1
B
x − 3
C
(x − 3)^2
Dx + E x^2 + 5 Each linear term in the denominator on the left gets a constant above it on the right; the squared linear factor (x − 3) on the left appears twice on the right, once to the second power. Each irreducible quadratic term on the left gets a linear term (Dx + E here) above it on the right.
- Trigonometric Substitutions: some suggested substitutions and useful formulae follow.
Radical Form
a^2 − x^2
a^2 + x^2
x^2 − a^2 Substitution x = a sin t x = a tan t x = a sec t
sin^2 x + cos^2 x = 1 tan^2 x + 1 = sec^2 x sin^2 x =
cos(2x) 2
cos^2 x =
cos(2x) 2 sin(2x) = 2 sin x cos x
- Powers of Trigonometric Functions: here are some strategies for dealing with these.
∫ sinm^ x cosn^ x dx Possible Strategy Identity to Use m odd Break off one factor of sin x and substitute u = cos x. sin^2 x = 1 − cos^2 x n odd Break off one factor of cos x and substitute u = sin x. cos^2 x = 1 − sin^2 x m, n even Use sin^2 x + cos^2 x = 1 to reduce to only powers of sin x sin^2 x =
cos(2x) 2 or only powers of cos x, then use table of integrals #39-42 cos^2 x =
cos(2x) 2 or identities shown to right of this box. ∫ tanm^ x secn^ x dx Possible Strategy Identity to Use m odd Break off one factor of sec x tan x and substitute u = sec x. tan^2 x = sec^2 x − 1 n even Break off one factor of sec^2 x and substitute u = tan x. sec^2 x = tan^2 x + 1 m even, n odd Use identity at right to reduce to powers of sec x alone. tan^2 x = sec^2 x − 1 Then use table of integrals #51.
Useful Trigonometric Derivatives d dx
sin x = cos x d dx
cos x = − sin x d dx
tan x = sec^2 x d dx
sec x = sec x tan x
- Improper integrals: look for ∞ as one of the limits of integration; look for functions that have a vertical asymptote in the interval of integration. It may be useful to know the following limits. lim x→∞ ex^ = ∞
lim x→∞ e−x^ = 0 Note: this is the same as lim x→−∞ ex
lim x→∞ 1 /x = 0 Note: the answer is the same for lim x→∞ 1 /x^2 and similar functions
lim x→ 0 +^
1 /x = ∞ Note: the answer is the same for lim x→ 0 +^
1 /x^2 and similar functions
lim x→∞ ln x = ∞ lim x→ 0 +
ln x = −∞
lim x→∞ arctan x = π/ 2
- Evaluate the following.
(a) Let u = sin x, so du = cos x dx. ∫ sin^6 x cos^3 x dx =
sin^6 x(1 − sin^2 x) cos x dx Use cos^2 x = 1 − sin^2 x.
u^6 (1 − u^2 ) du
(u^6 − u^8 ) du
u^7 7
u^9 9
+ C
sin^7 x 7
sin^9 x 9
+ C
(b) Let x = 5 sin t, so dx = 5 cos t dt. Note that this means t = arcsin(x/5). ∫ (^) √ 25 − x^2 dx =
25 − 25 sin^2 t · 5 cos t dt
1 − sin^2 t · 5 cos t dt Use 1 − sin^2 t = cos^2 t.
cos^2 t · cos t dt
cos^2 t dt
[
t 2
sin(2t) 4
]
cos(2t) 2
[
arcsin(x/5) 2
2 sin t cos t 4
]
- C Use sin(2t) = 2 sin t cos t.
[
arcsin(x/5) 2
x 5
25 − x^2 5
]
x 5
and cos t =
25 − x^2 5
[
arcsin(x/5) 2
x
25 − x^2 50
]
+ C
(f) This integral is improper at x = 1 because the integrand has a vertical asymptote there. ∫ (^3)
1
x − 1
dx = lim t→ 1 +
t
x − 1
dx
= lim t→ 1 +
ln |x − 1 |
∣^3
t = lim t→ 1 +
[ln | 3 − 1 | − ln |t − 1 |]
Since lim t→ 1 +
(− ln |t − 1 |) = ∞, this integral diverges (to ∞).
- Find the second-degree Taylor polynomial for f(x) =
x centered at x = 100.
f(x) = x^1 /^2 f(100) = 10
f′^ (x) =
x−^1 /^2 =
x
f′(100) =
f′′(x) =
x−^3 /^2 =
4 x^3 /^2
f′′(100) =
4 · 1003 /^2
P 2 (x) = f(100) + f′(100)(x − 100) +
f′′(100) 2!
(x − 100)^2
x − 100 20
(x − 100)^2 8000
- What is the maximum possible error that can occur in your Taylor approximation from the previous problem on the interval [100, 110]?
We know that |f(x) − Pn(x)| ≤ Kn+ (n + 1)!
|x − x 0 |n+1.
In this case, n = 2, x 0 = 100, and x = 110 (the farthest from x 0 that we are considering).
K 3 = max of |f′′′(x)| on [100, 110] = max of |
8 x^5 /^2
| on [100, 110] =
8 · 1005 /^2
Putting this all together, we have |f(x) − P 3 (x)| ≤
3 800 , 000 3!
| 110 − 100 |^3 =
- Use comparisons to show whether each of the following converges or diverges. If an integral converges, also give a good upper bound for its value.
(a)
1
6 + cos x x^0.^99 dx
For all x ≥ 1, we have
6 + cos x x^0.^99
x^0.^99
x^0.^99 because the minimum value of cos x is −1.
Since
1
5 dx x^0.^99
diverges (compute yourself or notice that p = 0. 99 < 1), we know that the integral in question must diverge too.
(b)
1
4 x^3 − 2 x^2 2 x^4 + x^5 + 1
dx
For all x ≥ 1, we have 4 x^3 − 2 x^2 2 x^4 + x^5 + 1
4 x^3 x^5
x^2
. (We’ve made the denominator smaller and the numerator larger, so the new fraction is larger.)
1
dx x^2
= 4 lim t→∞
∫ (^) t
1
dx x^2
= 4 lim t→∞
x
t 1 = 4 lim t→∞
[
t
]
= 4[0 − (−1)]
Therefore, the original integral in question must converge to a value less than 4.
- The lifespan of a certain type of light bulb is normally distributed with a mean of 2000 hours and a standard deviation of 300 hours. Write an integral equal to the fraction of bulbs that last between 1700 and 2600 hours.
We use the normal density function f(x) =
2 πs
e−^
(x−m)^2 2 s^2.
The mean is m = 2000 and the standard deviation is s = 300.
So, the fraction of bulbs that last between 1700 and 2600 hours is
2 π · 300
1700
e−^
(x−2000)^2 2 · (^3002).
(Notice that this is exactly one standard deviation below the mean and two standard deviations above the mean. Evaluating by numerical integration or by the Z-score table in the book gives a fraction of about 0.8186 or 81.86%.)
