Antiderivatives, Integration by Substitution - Lecture Notes | MATH 1451, Study notes of Mathematics

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5.1 Antiderivatives

Use the rules on the back cover table, not page 328. The rules on page 328 do not use the correct notation. The rules on the back cover do use the correct notation.

Definition: F is an antiderivative for f provided F ′^ = f. This is also called an indefinite

integral :

∫

f (x)dx = F (x) + C

Example: If f (x) = 2x, then F (x) = x^2 and G(x) = x^2 + 3 are both antiderivatives of f.

∗ Antiderivatives are unique up to a constant.

Antiderivative Rules - “Undo” differentiation. Think about what happens when you take a derivative. Now reverse the process. As antiderivatives “undo” differentiation, all the procedural rules we had for derivatives still apply:

Constant multiple rule:

∫

cf (u)du = c

∫

f (u)du

Sum/Difference rules:

∫

[f (u) ± g(u)] du =

∫

f (u)du ±

∫

g(u)du

Linearity rules:

∫

[af (u) ± g(u)] du = a

∫

f (u)du ± b

∫

g(u)du Here are two of the antiderivative rules. (Please see the back cover of your text book for the rest.)

Log Rule

∫

x−^1 dx =

∫

x

dx = ln |x| + C

Power Rule

∫

xndx =

n + 1

xn+1^ + C, n 6 = − 1

Examples

∫ (

x^2 + 4x^3

)

dx

∫

x

dx

∫

√

xdx

∫

x^2 + 4

√

x + 7 x^2

dx

∫

cos xdx

∫

sin xdx

∫

sec (t) tan (t)dt

∫

√

1 − x^2

dx

Slope Fields - Since a derivative is a slope, finding an antiderivative is like drawing the graph given the slope.

Ex: f ′(x) = 2x (slopes)

If we want an antiderivative through (1, 4), we imagine dropping a pebble in and letting the current carry it. ∫ 2 xdx = x^2 + C

5.5 Integration by Substitution

U-Substitution - for “undoing” chain rule ∫

2 cos(2x)dx = sin (2x) + C

∫

cos(2x)dx =

Method 1∫

cos(2x)dx

Method 2 - doesn’t solve for dx. Instead, we multiply by a special 1 ∫

cos(2x)dx

Examples

∫

x

√

x^2 + 3dx

∫

e^4 x

(^3) + 2 x^2 dx

∫

sec^2 (3x − 1)dx

∫

sec^2 x (tan (x) + 3)^4 dx

∫

sec^2 (3x)(tan (3x) + 3)^4 dx

∫

1 + x^2

dx

∫

ex 1 + e^2 x^

dx

∫

e

√t √ t(e

√ t (^) + 1)dt

∫

tan xdx

∫

x

√

3 x − 1 dx

5.2 Area Under a Curve

I. Area

Question: How can we find the area of a curved region? For example, the area of a rectangle is the length ∗ width.

But was about?

Estimate the area under f (x) = x^2 + 3 over [0, 2].

Idea: Use n = 4 rectangles to estimate the area.

In general: If f is continuous and f (x) ≥ 0 throughout [a, b], then we can approximate the area under the curve by summing rectangles.

A ≈ f (a + 4 x) 4 x + f (a + 2 4 x) 4 x + · · · + f (a + n 4 x) 4 x where 4 x =

b − a n

.

We can get a better approximation by summing the area of more (and smaller) rectangles. This means 4 x has to get small.

So the area under the curve is (not approximated by):

A = lim 4 x→ 0 [f (a + 4 x) + f (a + 2 4 x) + · · · + f (a + n 4 x)] 4 x where 4 x =

b − a n

.

Ex: Approximate the area between the x-axis and f (x) = 4x − 1 on [0, 1] with n = 4, using right endpoints:

The more rectangles and sums we do, the better notation we’ll need.

II. Sigma Notation

∑^ n

k=

ak = a 1 + a 2 + a 3 + · · · + an, k is an arbitrary index

Connection to area: Let Sn be the sum of the first n rectangles. Then

Sn = [f (a + 4 x) + f (a + 2 4 x) + · · · + f (a + n 4 x)] 4 x

=

∑^ n

k=

f (a + k 4 x) 4 x, where 4 x =

b − a n

Then A = lim n→∞

∑^ n

k=

f (a + k 4 x) 4 x

Rules for Summation (Theorem 5.4, pg 339)

For any numbers c & d, and positive integers m & n

1. Constant

∑^ n

k=

c = c︸ + c +︷︷ · · · + (^) ︸c n terms

= nc

2. Sum

∑^ n

k=

(ak + bk) =

∑^ n

k=

ak +

∑^ n

k=

bk

3. Scalar Multiplier

∑^ n

k=

cak = c

∑^ n

k=

ak

4. Linearity:

∑^ n

k=

(cak + dbk) =

∑^ n

k=

cak +

∑^ n

k=

dbk

5. Subtotal: If 1 < m < n,

∑^ n

k=

ak =

∑^ m

k=

ak +

∑^ n

k=m+

ak

6. Dominance: If ak ≤ bk,

∑^ n

k=

ak ≤

∑^ n

k=

bk

Summation Formulas(pg 340)

∑^ n

k=

1 = n

∑^ n

k=

k = 1 + 2 + · · · + n =

n(n + 1) 2

∑^ n

k=

k^2 = 1^2 + 2^2 + 3^2 + · · · + n^2 =

n(n + 1)(2n + 1) 6

∑^ n

k=

k^3 = 1^3 + 2^3 + 3^3 + · · · + n^3 =

n^2 (n + 1)^2 4

Ex: Find the area under f (x) = x^2 on [0, 1].

Example continued.

5.3/5.4 Definite Integrals

I. Area/Definite Integral

Recall: We can find the area under a curve by summing rectangles. The more rectangles we use, the more accurate the calculation.

A = lim n→∞

∑^ n

k=

f (a + k 4 x) 4 x, with 4 x =

b − a n

gives area under f (x) on [a, b]

Definite Integral:

∫ (^) b

a

f (x)dx = lim n→∞

∑^ n

k=

f (a + k 4 x) 4 x

II. Fundamental Theorem of Calculs(Part 1)

If f is continuous on [a, b], then

∫ (^) b

a

f (x)dx = F (b) − F (a, ) where F is an antiderivative of

f.

Ex: Find the area under f (x) = x^2 + 3 on [0, 2].

III. Properties of the Definite Integral

∫ (^) a

a

f (x)dx = 0

∫ (^) b

a

f (x)dx = −

∫ (^) b

a

f (x)dx

3. If f (x) ≤ g(x),

∫ (^) b

a

f (x)dx ≤

∫ (^) b

a

g(x)dx (Dominance)

∫ (^) b

a

[rf (x) + sg(x)]dx = r

∫ (^) b

a

f (x) + s

∫ (^) b

a

g(x)dx (Linearity)

5. If a < c < b,

∫ (^) b

a

f (x) =

∫ (^) c

a

f (x)dx +

∫ (^) b

c

f (x)dx (Subdivision)

Ex: Find the area ‘beneath’ f (x) = sin(x) on [0, 2 π].

Ex: If

∫ 7

1

f (x)dx = 3 and

∫ 7

4

f (x)dx = 10, find

∫ 4

1

f (x)dx.

IV. Fundamental Theorem of Calculus(Part 2)

If f is continuous on [a, b], then

d dx

∫ (^) x

a

f (t)dt = f (x).

Note: Not every continuous function has a derivative, but every continuous function has an antiderivative.

“Proof”:

Ex:

d dx

∫ (^) x

a

t^2 − 4 tdt

Ex:

d dx

∫ 3

x

sec^2 u du

Ex:

d dx

∫ 0

x^2 +4x

t

dt

Note:

∫ (^) b

a

f (x) dx = number

∫ (^) x

a

f (t) dt = a function of x

∫ f (x) dx = a family of functions (of x)

5.7 Mean Value Theorem for Integrals

I. Mean Value Theorem (for integrals)

Area under curve:

∫ (^) b

a

f (x) dx

Area of box: f (c)(b − a)

If we pick c carefully, we can make these areas the same.

Mean Value Theorem: If f is continuous on [a, b], then there is some c in (a, b) such that:

∫ (^) b

a

f (x) dx = f (c)(b − a)

Ex: Find the value of c guaranteed by The Mean Value Theorem for integrals for

f (x) =

x^3

on [1, 3].

II. Average Value

If f is continuous on [a, b] the average value (AV) of f on this interval is:

AV =

b − a

∫ (^) b

a

f (x) dx

Ex: Find the average value of f (x) =

√ 3

1 − x on [− 7 , 0].

Ex: The temperature (in degrees Fahrenheight) varies according to:

F (t) = 6. 44 t − 0. 23 t^2 + 30.

Find the average daily temperature if t is the time of day (in hours) from midnight.

5.8 Numerical Integration

The Fundamental Theorem of Calculus (part 1) tells us how to calculate

∫ (^) b

a

f (x) dx

if we can find an antiderivative. But what if we can’t?

For example, f (x) = e−x 2 has no simple antiderivative: ∫ e−x

2 dx =

√

π 2

erf (x) + C

Instead of using an antiderivative, we can only approximate the definite integral.

When using rectangles, we increase accuracy by making n larger (more and smaller rect- angles). An alternative to this is to change the shape (then we could use a smaller n and still have better accuracy).

I. Trapezoid Rule - averages left and right endpoints

Atrap =

(f (x 1 ) + f (x 2 )) 4 x

∫ (^) b

a

f (x)dx ≈

[f (x 0 ) + f (x 1 )] 4 x +

[f (x 1 ) + f (x 2 )] 4 x + · · · +

[f (xn− 1 ) + f (xn)] 4 x

4 x =

b − a n

and xk = a + k 4 x

Ex: Approximate

∫ 2

0

x^3 + 3 dx using the trapezoid rule and n = 4.

Simpson’s Rule - parabola tops

Simpson’s Rule:

∫ (^) b

a

f (x)dx ≈

[f (x 0 )+4f (x 1 )+2f (x 2 )+4f (x 3 )+2f (x 4 )+· · ·+2f (xn− 2 )+4f (xn− 1 )+f (xn)] 4 x

Ex: Approximate

∫ 2

0

x^3 + 3 dx using n = 4 intervals and Simpson’s Rule.