# Recall from Math 141 Integration Basics Fundamental Theorem of

## Transcription

Recall from Math 141 Integration Basics Fundamental Theorem of
```Recall from Math 141
Integration Basics
Fundamental Theorem of Calculus (FTC)
Let f : [a, b] → R be a continuous function.
Let F : [a, b] → R be a function.
• If F is an antiderivative of f on [a, b] (i.e. F 0 (x) = f (x) for each x ∈ [a, b]), then
Z b
Z b
f (x) dx ≡
F 0 (x) dx = F (b) − F (a) .
a
a
Rx
• If F (x) = a f (t) dt for each x ∈ [a, b], then F is an antiderivative of f on [a, b], i.e.
Z x
0
F (x) ≡ Dx
f (t) dt = f (x) .
a
Basic Differentiation Rules
If the functions y = f (x) and y = g(x) are differentiable at x and a and b are constants, then:
(1) Dx [af (x) + bg(x)] = af 0 (x) + bg 0 (x)
(2) Dx h[f (x)i · g(x)] = f 0 (x) · g(x) + f (x) · g 0 (x)
0
(x)
− f (x) · g 0 (x)
(3) Dx fg(x)
= f (x) · g(x)[g(x)]
provided g(x) 6= 0 .
2
If f is differentiable at x and g is differentiable at f (x), then:
(4) Dx [g(f (x))] = g 0 (f (x)) f 0 (x) .
On this handout, a (∗) means that you do not have to memorize the formula
but should be able to use the formula if you are given it.
Hyperbolic Trig Functions1
ex + e−x
2
(∗) sinh x
tanh x =
cosh x
(∗)
(∗)
cosh x =
sinh x =
(∗)
coth x =
derivatives
cosh x
sinh x
ex − e−x
2
(∗)
cosh2 x − sinh2 x = 1
(∗)
sech x =
FTC
=⇒
du
dx
du
(∗)
Dx sinh u = cosh u
dx
1
du
(∗)
Dx sinh−1 u = √
2
1 + u dx
1
du
(∗)
Dx cosh−1 u = √
2
u>1
u − 1 dx
Dx cosh u = sinh u
1Hyperbolic
(∗)
csch x =
1
sinh x
integrals
Z
(∗)
1
cosh x
(∗)
sinh u du = cosh u + C
Z
(∗)
cosh u du = sinh u + C
Z
du
(∗)
−1 u
√
= sinh
+C
a
a2 + u2 a>0
Z
du
(∗)
−1 u
√
= cosh
+C
a
u2 − a2 u>a>0
Trig Functions are covered in §7.3, pages 439–447.
16.08.17 (yr.mn.dy)
Page 1 of 4
Integration
Recall from Math 141
Integration Basics
Basic Integral Formulas
FTC
=⇒
derivatives
Dx un = nun−1
integrals
Z
du
dx
du
Dx e = e
dx
u6=0 1 du
Dx ln |u| =
u dx
du
Dx au = au ln a
0 < a 6= 1
dx
1 du
u6=0 1
Dx loga |u| =
0 < a 6= 1
u ln a dx
du
Dx sin u = cos u
dx
du
Dx tan u = sec2 u
dx
du
Dx sec u = sec u tan u
dx
du
Dx cos u = − sin u
dx
du
Dx cot u = − csc2 u
dx
du
Dx csc u = − csc u cot u
dx
1
du
Dx sin−1 u = √
= −Dx cos−1 u
2
1 − u dx
1
du
Dx tan−1 u =
= −Dx cot−1 u
2
1 + u dx
1
du
√
Dx sec−1 u =
= −Dx csc−1 u
2
|u| u − 1 dx
more integrals
u
u
Z
un du
n6=−1
=
un+1
+ C
n+1
eu du = eu + C
Z
du u6=0
= ln |u| + C
u
Z
au
au du =
+ C
ln a
Z
cos u du = sin u + C
Z
sec2 u du = tan u + C
Z
sec u tan u du = sec u + C
Z
sin u du = − cos u + C
Z
csc2 u du = − cot u + C
Z
csc u cot u du = − csc u + C
Z
du
u
a>0
√
+ C
= sin−1
2
2
a
a −u
Z
du
u
a>0 1
tan−1
+ C
=
2
2
a +u
a
a
Z
du
|u|
a>0 1
√
=
sec−1
+C
2
2
a
a
u u −a
Z
tan u du
=
− ln |cos u| + C
=
ln |sec u| + C
cot u du
=
ln |sin u| + C
=
− ln |csc u| + C
sec u du
=
ln |sec u + tan u| + C
=
− ln |sec u − tan u| + C
csc u du
=
− ln |csc u + cot u| + C
=
ln |csc u − cot u| + C
Z
Z
Z
16.08.17 (yr.mn.dy)
Page 2 of 4
Integration
Recall from Math 141
Integration Basics
Generalized Exponential y = bx and Logarithmic y = logb x Functions
with base b where b > 0 but b 6= 1. Also 0 < a 6= 1.
ln ≡ loge
f (x) = bx ≡ ex ln b
: (−∞, ∞) → (0, ∞)
g(x) = logb x ≡ the inverse of the fn. f (x) = b
y = logb x
⇐⇒
(loga b)(logb c) = loga c
x
: (0, ∞) → (−∞, ∞)
x = by
ln x
loga x =
ln a
=⇒
x, y > 0 & r ∈ R
x, y ∈ R & r ∈ R
blogb x = x
logb (bx ) = x
logb 1 = 0
b0 = 1
logb (xy) = logb x + logb y
x
= logb x − logb y
logb
y
logb (xr ) = r(logb x)
bx by = bx+y
bx
= bx−y
by
(bx )r = bxr
x
x x
(ab) = a b
and
a x
b
ax
= x
b
Basic Trig
hypotenuse
cos θ =
hyp
sin θ =
opp
hyp
tan θ =
opp
cot x =
cos x
sin x
sec x =
1
cos x
csc x =
1
sin x
opposite
θ
tan x =
sin x
cos x
Basic Inverse Trig Functions
y = sin θ
⇔
θ = sin−1 y
where
−1≤y ≤1
and
y = cos θ
⇔
θ = cos−1 y
where
−1≤y ≤1
and
y = tan θ
⇔
θ = tan−1 y
where
y∈R
and
y = cot θ
⇔
θ = cot−1 y
where
y∈R
and
y = sec θ
⇔
θ = sec−1 y
where
|y| ≥ 1
and
y = csc θ
⇔
θ = csc−1 y
where
|y| ≥ 1
and
16.08.17 (yr.mn.dy)
Page 3 of 4
−π
π
≤θ≤
2
2
0≤θ≤π
−π
π
<θ<
2
2
0<θ<π
π
0 ≤ θ ≤ π , θ 6=
2
−π
π
≤ θ ≤ , θ 6= 0
2
2
Integration
To come in Math 142
Integration Basics
Our Math 142 Course Homepage (CH) is
http://people.math.sc.edu/girardi/w142.html
Integration by Parts
Z
Z
u dv = uv − v du
Trig Identities useful for Integration
Double-Angle Formulas:
cos(2x) = cos2 x − sin2 x
sin(2x) = 2 sin x cos x
Half-Angle Formulas:
cos2 x =
1 + cos(2x)
2
sin2 x =
1 − cos(2x)
2
(∗)
(∗)
cos(s + t) = cos s cos t − sin s sin t
sin(s + t) = sin s cos t + cos s sin t
(∗)
(∗)
cos(s − t) = cos s cos t + sin s sin t
sin(s − t) = sin s cos t − cos s sin t
Trig Substitution
if integrand involves
a2 − u 2
a2 + u 2
u 2 − a2
then make the substitution
u
u = a sin θ ! θ = sin−1
a
−1 u
u = a tan θ ! θ = tan
a
−1 u
u = a sec θ ! θ = sec
a
restriction on θ
−π
π
≤θ≤
2
2
−π
π
<θ<
2
2
π
0 ≤ θ ≤ π , θ 6=
2
16.08.17 (yr.mn.dy)
Page 4 of 4
Integration
```

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