FTC Review;The Method of Substitution

February 4, 2004

The Definite Integral as Area

Let f be a continuous function defined on the interval [a, b]. The definite integral of f from a to b, denoted by

represents the total signed area of the region bounded by y = f (x), the vertical lines x = a and x = b, and the x-axis.

dxxfb

a

Properties of Definite Integrals

1.

2.

3.

4.

5.

6.

7.

Let f and g be continuous functions defined on the interval [a, b]. Furthermore, let c and k be constants such that a < c < b. Then…

0a

a

dxxf

b

a

b

a

dxxfkdxxfk

b

a

b

a

b

a

dxxgdxxfdxxgxf

b

a

b

a

b

a

dxxgdxxfdxxgxf

b

a

a

b

dxxfdxxf

b

a

b

a

dxxgdxxfbaxgxf . then ,on If

b

c

c

a

b

a

dxxfdxxfdxxf

The Fundamental Theorem of Calculus

Let f be a continuous function defined on [a, b], and let F be any antiderivative of f. Then

).()( aFbFdxxfb

a

Keeping It Straight

Definite IntegralRepresents a real number (a signed area).

Area FunctionRepresents a single antiderivative of f.

Indefinite IntegralRepresents the entire family of antiderivatives of f.

b

a

dxxf

x

a

dttf

dxxf

Substitution Rule for Indefinite Integrals

duufdxxgxgf

xgu

then, If

Implementing the Substitution Rule

1. Choose u.

2. Differentiate u w.r.t. x and solve for du.

3. Substitute u and du into the old integral involving x to form a new integral involving only u.

4. Antidifferentiate with respect to u.

5. Re-substitute to find the antiderivative as a function of x.

Two Special Forms

Cxgdxxg

xg

Cedxexg xgxg

ln

Substitution Rule for Definite Integrals

bg

ag

b

a

duufdxxgxgf

xgu then, If

Implementing the Substitution Rule(Definite Integrals)

1. Choose u = g(x).

2. Differentiate u w.r.t. x and solve for du.

3. Substitute u and du into the old integral involving x, as well as converting endpoints from a and b to g(a) and g(b).

4. Antidifferentiate with respect to u and evaluate at the new endpoints.

4

0

2 12 dxxx 17

1

duu

Arcsine (Inverse Sine Function)

For x in [-1, 1], y = arcsin x is defined by the conditions

i) x = sin y and

ii) –/2 y /2.

In words, arcsin x is the angle between –/2 and /2 whose sine is x.