INTEGRALS

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5. INTEGRALS. INTEGRALS. Indefinite Integrals. INDEFINITE INTEGRAL. The notation ∫ f ( x ) dx is traditionally used for an antiderivative of f and is called an indefinite integral. Thus, ∫ f ( x ) dx = F ( x ) means F’ ( x ) = f ( x ). INDEFINITE INTEGRALS.

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5

INTEGRALS

INTEGRALS

Indefinite Integrals

INDEFINITE INTEGRAL

The notation ∫ f(x) dx is traditionally used for an antiderivative of f and is called an indefinite integral.

Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)

INDEFINITE INTEGRALS

For example, we can write

• Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).
INDEFINITE INTEGRALS

Any formula can be verified by differentiating the function on the right side and obtaining the integrand.

For instance,

INDEFINITE INTEGRALS

Thus, we write with the understanding that it is valid on the interval (0, ∞) or on the interval (-∞, 0).

INDEFINITE INTEGRALS

This is true despite the fact that the general antiderivative of the function f(x) = 1/x2, x ≠ 0, is:

INDEFINITE INTEGRALS

Example 1

Find the general indefinite integral ∫ (10x4 – 2 sec2x) dx

• Using our convention and Table 1, we have: ∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx= 10(x5/5) – 2 tan x + C= 2x5 – 2 tan x + C
• You should check this answer by differentiating it.
INDEFINITE INTEGRALS

Example 2

Evaluate

• This indefinite integral isn’t immediately apparent in Table 1.
• So, we use trigonometric identities to rewrite the function before integrating:

INTEGRALS

The Substitution Rule

In this section, we will learn:

To substitute a new variable in place of an existing

expression in a function, making integration easier.

INTRODUCTION

Equation 1

• Our antidifferentiation formulas don’t tell us how to evaluate integrals such as
INTRODUCTION
• To find this integral, we use the problem-solving strategy of introducing something extra.
• The ‘something extra’ is a new variable.
• We change from the variable x to a new variable u.
INTRODUCTION
• Suppose we let u = 1 + x2 be the quantity under the root sign in the integral below,
• Then, the differential of u is du = 2xdx.
INTRODUCTION
• Notice that, if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in
INTRODUCTION

Equation 2

• So, formally, without justifying our calculation, we could write:
INTRODUCTION
• However, now we can check that we have the correct answer by using the Chain Rule to differentiate
INTRODUCTION
• In general, this method works whenever we have an integral that we can write in the form ∫f(g(x))g’(x) dx
INTRODUCTION

Equation 3

• Observe that, if F’ = f, then
• ∫F’(g(x))g’(x) dx = F(g(x))+ C
• because, by the Chain Rule,
INTRODUCTION
• If we make the ‘change of variable’ or ‘substitution’ u = g(x), we have:
INTRODUCTION
• Writing F’ = f, we get: ∫f(g(x))g’(x) dx = ∫f(u) du
• Thus, we have proved the following rule.
SUBSTITUTION RULE

Equation 4

• If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫f(g(x))g’(x) dx = ∫f(u) du
SUBSTITUTION RULE
• Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation.
• Notice also that, if u = g(x), then du = g’(x) dx.
• So,a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.
SUBSTITUTION RULE
• Thus, the Substitution Rule says:
• It is permissible to operate withdx and du after integral signs as if they were differentials.
SUBSTITUTION RULE

Example 1

• Find ∫x3 cos(x4 + 2) dx
• We make the substitution u = x4 + 2.
• This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.
SUBSTITUTION RULE

Example 1

• Thus, using x3dx = du/4 and the Substitution Rule, we have:
SUBSTITUTION RULE
• The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral.
• This is accomplished by changing from the original variable x to a new variable u that is a function of x.
• Thus, in Example 1, we replaced the integral ∫ x3cos(x4 + 2) dxby the simpler integral ¼ ∫ cos u du.
SUBSTITUTION RULE
• The main challenge in using the rule is to think of an appropriate substitution.
• You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor.
• This was the case in Example 1.
SUBSTITUTION RULE
• If that is not possible, try choosing u to be some complicated part of the integrand—perhaps the inner function in a composite function.
SUBSTITUTION RULE
• Finding the right substitution is a bit of an art.
• It’s not unusual to guess wrong.
• If your first guess doesn’t work, try another substitution.
SUBSTITUTION RULE

E. g. 2—Solution 1

• Evaluate
• Let u = 2x + 1.
• Then, du = 2 dx.
• So, dx = du/2.
SUBSTITUTION RULE

E. g. 2—Solution 1

• Thus, the rule gives:
SUBSTITUTION RULE

Example 3

• Find
• Let u = 1 – 4x2.
• Then, du = -8x dx.
• So, x dx = -1/8 du and
SUBSTITUTION RULE

Example 4

• Calculate ∫ cos 5xdx
• If we let u = 5x, then du = 5 dx.
• So, dx = 1/5 du.
• Therefore,