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CHAPTER 6: INTEGRAL CALCULUS

In Chapter 5, we discussed about the derivatives of functions f, f’ . In this chapter we will discuss how to go from f’ to f.

6.1 ANTI- DIFFERENTIATION

Given f, we find f’ : DIFFERENTIATION

Given f’, we find f : ANTI- DIFFERENTIATION

Anti-differentiation is the inverse of differentiation

Example:

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f(x) =x2

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⇒

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′ f (x) =2x

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So antiderivatives of ′ f (x) =2x is x2.

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Antiderivatives of a function g is called an integration of g.

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g(x) dx∫

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In previous example, we have 2x dx =x2∫ .

Another example:

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ddx

(2x) =2

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⇒

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2 dx∫ =?

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ddx

(2x +3) =2

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⇒

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2 dx∫ =?

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ddx

(2x −5) =2

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⇒

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2 dx∫ =?

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ddx

(2x +c) =2

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⇒

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2 dx∫ =?

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In general:

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f(x)dx =F(x)+c ∫

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→∫ integral sign- indefinite integral.

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f(x) → the integrand.

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dx → the variable of integration.

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c → constant of integration.

Next we want to see the basic rules of integration. All the basic rules of integration can be easily be derived by using the inverse relationship between differentiation and integration.

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BASIC RULES OF INTEGRATION

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Have ddx

(c) =0, where c∈ R ⇒ 0 dx =c.∫

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0dx =c∫

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kdx=kx+c∫

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xn∫ dx =xn+1

n +1+c (n ≠1)

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sinx ∫ dx =−cosx +c

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cosx ∫ dx =sinx +c

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1x∫ dx =lnx +c

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ex∫ dx =ex +c

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kf(x)∫ dx =k f(x)dx∫

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f(x)±g(x)[ ]∫ dx = f(x)dx + g(x)dx∫∫

Note that:

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f(x)• g(x)[ ]∫ dx ≠ f(x)dx • g(x)dx∫∫

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f(x)g(x)∫ dx ≠

f(x)∫ dxg(x)dx∫

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Example 1: Indefinite Integrals

Find the following integral

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a) (5x +7)dx∫

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b) (3x + x2∫ )dx

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c) (sinx +3cosx)dx∫

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d ) 3t−2t2

⎛ ⎝ ⎜

⎞ ⎠ ⎟∫

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6.2 Integration of Linear Function With Power

Our goal in this section is to integrate a function of the form

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(ax+ b)n

The expression in the bracket must be linear.

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(ax+b)n∫ =

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(ax+ b)n+1

(n +1)

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÷ddx

(ax+ b)

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=(ax+ b)n+1

a(n +1)

Example 1: Integration of linear with power

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a) (3x +2)3dx∫

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b) 2x +3 dx∫

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c) 4

4 −x dx∫

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d ) 6

2x +2 dx∫

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6.3 SUBSTITUTION RULE

Our goal in this section is to expand our ability to compute antiderivatives through a useful technique called integration by substitution. As we have already seen, we need the chain rule to compute derivatives of many functions. Integration by substitution gives us process for helping to recognize when an integrand is the result of a chain rule derivative.

Integration by substitution consists of the following general steps:

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• Choose an expression for u.A common choice is the innermost expression or “inside term” of composition of functions.

Example:

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(x3∫ +5)100(3x2 )dx

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Choose: u =(x3 +5) Why?

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Since: dudx

=3x2

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⇒ du =3x2dx

• Compute du/dx and find du.• Replace all terms in the original integrand with expressions involving u and du.

Example:

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(x3∫ +5)100(3x2 )dx

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= u∫ 100du

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• Evaluate the resulting u integral as usual. If you can’t evaluate the integral, you may need to try a different choice of u.

Example:

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(x3∫ +5)100(3x2 )dx

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= u100∫ du

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=u101

101+c

• Replace each occurrence of u in the antiderivatives with the corresponding expression in x.

Example:

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(x3∫ +5)100(3x2 )dx

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=u101

101+c

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=(x3 +5)101

101+c

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Example 1: Integration by Substitution

Evaluate the following integration

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a) xcosx2 dx∫

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b) te(t2 +1)∫ dt

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c) x3∫ x4 +5 dx

Solution:

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6.4 DEFINITE INTEGRALS

If f is continuous on the closed interval [a, b], then f is integrable on [a, b]. Thus the integral of f from x=a to x=b denoted by

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f(x)dxa

b

∫is called a definite integral and is defined by

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f(x)dxa

b

∫ =F(x)a

b=F(b)−F(a)

where a is the lower limit of the integral and b is the upper limit of the integral.

NOTE: Need to integrate as usual using the rules, then substitute the upper and lower limit.

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Example 1: Definite Integral

Evaluate the following integration.

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a) (x3

-1

3

∫ +5x)dx

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b) dx5−x1

3

∫

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c) 1

3+2x0

1

∫ dx