Indefinite Integrals

Indefinite Integrals

If F(x) is a function whose derivative F'(x) = f(x) on certain interval of the x-axis, then F(x) is called the anti-derivative of indefinite integral f(x). When we integrate the differential of a function we get that function plus an arbitrary constant. In symbols we write

where the symbol symbol , called called the integral integral sign, specifies the operation of integration integration upon f(x) dx; that is, we are to find a function whose derivative is f(x) or whose differential is f(x) dx. The dx tells us that the variable of integration is x. Integration Formulas

In these formulas, u and v denote differentiable functions of some independent variable (say x) and a, n, and C are constants.

1. The integral of the differential of a function u is u plus an arbitrary constant C (the definition of an integral).

2. The integral of a constant times the differential of the function. (A constant may be written before the integral sign but not a variable factor).

3. The integral of the sum of a finite number of differentials is the sum of their integrals.

4. If n is not equal to minus one, the integral of un du is obtained by adding one to the exponent and divided by the new exponent. This is called the General Power Formula .

1 - 3 Examples | Indefinite Integrals E

valuate the following integrals:

Example

1:

Example

2:

Example

3:

Solution to Example 1:

4. If n is not equal to minus one, the integral of un du is obtained by adding one to the exponent and divided by the new exponent. This is called the General Power Formula .



Example

1:

Example

2:

Example

3:


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This is the form . If we let multiplied by the differential integral can be evaluated as follows:

, then is raised to a power 4 and is of the function corresponding correspond ing to , the

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It should be pointed out that no integral can be evaluated directly unless it contains, in addition to the expression identified with , the exact differential of the function corresponding correspo nding to .


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valuate the following: following:

Example

4:

Example

5:

Example

6:

Solution to Example 4

This is not of the form because of the missing constant factor 3 in the integrand. Identifying , , then the differential . We must then insert 3 in the integrand and to compensate for it, we place the reciprocal 1/3 before the integral sign. This in effect multiplying by one does not affect the value of the function.

Let

, then

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Let u = 2x3 + 2x + 1 du = (6x2 + 2) dx = 2(3x2 + 1) dx n = -2/3

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If we let and , then . But there is no in the given integrand. It is easy to insert -4 in the integrand and offset this by placing -1/4 before the integral sign but nothing can be done about the missing factor . We therefore expand and integrate term by term.

answer

Definite Integral

The definite integral of f(x) is the difference between two values of the integral of f(x) for two distinct values of the variable x. If the integral of f(x) dx = F(x) + C, the definite integral is denoted by the symbol

The quantity F(b) - F(a) is called the definite integral of f(x) between the limits a and b or simply the definite integral from a to b. It is called the definite integral because the result involves neither x nor the constant C and therefore has a definite value. The numbers a and b are called the limits of integration, a being the lower limit and b the upper limit.

General

Properties of Definite Integral

1. The sign of the integral changes if the limits are interchanged.

2. The interval of integration may be broken up into any number of subintervals, and integrate over each interval separately.

3. The definite integral of a given integrand is independent of the variable of integration. Hence, it makes no difference what letter is used for the variable of integration.

Chapter 2 - Fundamental Integration The General Power Formula Logarithmic Functions Exponential Functions Trigonometric Functions Trigonometric Transformation Inverse Trigonometric Functions

Formulas

The General Power Formula |

Fundamental Integration Formulas

The General Power Formula as shown in Chapter 1 is in the form

Thus far integration has been confined to polynomial functions. Although the power formula was studied, our attention was necessarily limited to algebraic integrals, so that further work with power formula is needed. The power formula can be used to evaluate certain integrals involving powers of the trigonometric functions.

Example 01 | The General Power Formula Problem E

valuate

Solution

answer

Example 02 | The General Power Formula Problem E

valuate

.

Solution

answer

Example 03 | The General Power Formula Problem

Integrate

.

Solution

answer

Logarit hmic

Functions | Fundamental Integration Formulas

The limitation of the Power Formula , is when ; this makes the right side of the equation indeterminate. This is where the logarithmic function comes in, note that

, and we can recall that

. Thus,

The formula above involves a numerator which is the derivative of the denominator. The denominator represents any function involving any independent variable. The formula is meaningless when is negative, since the logarithms of negative numbers have not been defined. If we write so that , then we have

When negative numbers are involved, the formula should be considered in the form

The integral of any quotient whose numerator is the differential of the denominator is the logarithm of the denominator. Exponential Functions | Fundamental Integration Formulas

There are two basic formulas for the integration of exponential functions.

1. 2.

Where = function, say = constant (example: 3, , sin 30°, ¥7)

Trigonometric Functions | Fundamental Integration Formulas Basic

1. 2. 3. 4. 5. 6.

Formulas

Formulas Derived from Logarithmic Function

7. 8. 9. 10.

The six basic formulas for integration involving trigonometric functions are stated in terms of appropriate pairs of functions. An integral involving and , which the simple integration formula cannot be applied, we must put the integrand entirely in terms of and or in terms of and . Notice that these formulas are reverse formulas in Differential Calculus.

The formulas derived from trigonometric function can be traced as follows:

 Formula

 Formula

Inverse Trigonometric Functions | Fundamental Integration Formulas

In applying the formula (Example: Formula 1 below), it is important to note that the numerator du is the differential of the variable quantity u which appears squared inside the square root symbol. We mentally put the quantity under the radical into the form of the square of the constant minus the square of the variable.

1. 2. 3. Chapter 3 - Techniques of Integration

Integration by Parts Integration by Substitution Integration of Rational Fractions Change of Limits with Change of Variable

Integration by Parts | Techniques of Integration

When and are differentiable functions of , When this is integrated we have

or

.

The expression to be integrated must be separated into two parts, one part being and the other part, together with , being . The factor corresponding to must obviously contain the differential of the variable of integration.

Integration by Substitution | Techniques of Integration There are two types of substitution: algebraic substitution and trigonometric substitution.

Algebraic Substitution | Integration by Substitution In algebraic substitution we replace the variable of integration by a function of a new variable. A change in the variable on integration often reduces an integrand to an easier integrable form.

1 - 3 Examples | Algebraic Substitution E


Example

1:

Example

2:

Example

3:


Let

answer


Let


Let

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Trigonometric Substitution | Techniques

of Integration

Trigonometric substitution is employed to integrate expressions involving functions of (a2 í u2), (a2 + u2), and (u2 í a2) where "a" is a constant and "u" is any algebraic function. Substitutions

convert the respective functions to expressions in terms of trigonometric functions. The substitution is more useful but not limited to functions involving radicals.

Use the following suggestions:

When the integrand involves... y y y

(a2 í u2), try u = a sin  (a2 + u2), try u = a tan  (u2 í a2), try u = a sec 

The substitution may be represented geometrically by constructing a right triangle. Integration of Rational Fractions | Techniques of Integration Partial Fraction

Functions of x that can be expressed in the form P(x)/Q(x), where both P(x) and Q(x) are polynomials of x, is known as rational fraction. A rational fraction is known to be a proper fraction if the degree of P(x) is less than the degree of Q(x). Example of proper fraction is...

A rational fraction is said to be an improper fraction if the degree of P(x) is greater than or equal to the degree of Q(x). Examples are...

and

Improper fraction may be expressed as the sum of a polynomial and a proper fraction. For example:

Proper fraction such as can be expressed as the sum of partial fraction , provided that the denominator will f act orized .

Integration of any rational fraction depends essentially on the integration of a proper fraction by expressing it into a sum of partial fractions. There are four cases that may arise in dealing with integrand involving proper fraction. Chapter 4 - Applications of Integration Plane Areas Solids of Revolution

Plane Areas in Rectangular Coordinates | Applications of Integration

There are two methods for finding the area bounded by curves in rectangular coordinates. These are... 1. by using a horizontal element (called strip) of area, and 2. by using a vertical strip of area. The strip is in the form of a rectangle with area equal to length × width, with width equal to the differential element . To find the total area enclosed by specified curves, it is necessary to sum up a series of rectangles defined by the strip.

Using Horizontal Strip

From the figure, the area of the strip is , where . The total area can be found by running this strip starting from going to . Our formula for integration is...

Note that is the right end of the strip and is always on the curve the strip and is always on the curve . We therefore substitute to integration.

and

is the left end of and prior

Using Vertical Strip

We apply the same principle of using horizontal strip to the vertical strip. Consider the figure below.

The total area is...

Where = upper end of the strip = = lower end of the strip =

The steps in finding the area can be outlined as follows: 1. Sketch the curve 2. Decide what strip to use and define its limits 3. Apply the appropriate formula based on the strip then integrate.

Browse for some examples that illustrate the two methods. Example 1 | Plane Areas in Rectangular Coordinates Example

1

Find the area bounded by the curve y = 9 - x2 and the x-axis.

Solution Step 1: Sketch the curve.

 downward parabola, vertex at (0, 9), latus rectum = 1

The required area is symmetrical with respect to the y-axis, in this case, integrate the half of the area then double the result to get the total area. The use of symmetry will greatly simplify our solution most especially to curves in polar coordinates.

Using Horizontal Strip Step 2: Determine the limits of the strip.

The strip shown will start from

and end to

Step 3: Apply the appropriate formula then integrate.

Where = parabola = = y-axis =

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Using Vertical Strip Step 2: Determine the limits of the strip.

In this case the limits are not defined; we need to solve the points of intersection of the curves. when

,

. The strip will swipe from x = 0 to x = 3.

Step 3: Apply the appropriate formula then integrate.

Where = parabola = = x-axis =

ok!

Example 2 | Plane Areas in Rectangular Coordinates Example

2

Find the area bounded by the curve a2 y = x3, the x-axis and the line x = 2a.

Solution

when when points of intersection: (0, 0) and (2a, 8a)

Using vertical strip:

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3

Find the area bounded by the curve x = y2 + 2y and the line x = 3.

Solution

 rightward parabola, vertex at (-1, -1), LR = 1

when


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4

Solve the area bounded by the curve y = 4x - x2 and the lines x = -2 and y = 4.

Solution

 downward parabola, vertex at (2, 4), LR = 1

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5

Find the area between the curves 2x2 + 4x + y = 0 and y = 2x.

Solution

 downward parabola, vertex at (-1, 2), LR = 1/2

Points of intersection

when x = 0, y = 2(0) = 0 when x = -3, y = 2(-3) = -6 points of intersection: (0, 0) and (-3, -6)

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6

Find each of the two areas bounded by the curves y = x3 - 4x and y = x2 + 2x.

Solution

x2 + 2x + 1 = y + 1  upward parabola, vertex at (-1, -1), LR = 1

 (approximate - for graphing purposes only)

Maximum and minimum points of y =

3

x

- 4x

when x = 1.15, y = 1.153 - 4(1.15) = -3.08 when x = -1.15, y = (-1.15)3 - 4(-1.15) = 3.08 Maxima: (-1.15, 3.08) Minima: (1.15, -3.08)

Points of intersection:

when x = 0, y = 0 when x = 3, y = 3^3 - 4(3) = 15 when x = -2, y = (-2)^3 - 4(-2) y = 0 Points of intersection: (0, 0), (3, 15) and (-2, 0)

answer

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Plane Areas in Polar Coordinates | Applications of Integration

The fundamental equation for finding the area enclosed by a curve whose equation is in polar coordinates is...

Where 1 and 2 are the angles made by the bounding radii.

The formula above is based on a sector of a circle with radius r and central angle d. Note that r is a polar function or r = f(). See figure above. Example 1 | Plane Areas in Polar Coordinates Example

1

Find the area enclosed by r = 2a sin2 .

Solution

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Example 2 | Plane Areas in Polar Coordinates Example

2

Find the area bounded by the lemniscate of Bernoulli r 2 = a2 cos 2.

Solution

The curve is symmetrical with respect to the origin, and occurs only with values of  from -45° to 45° (-¼  to ¼ ).

The area in polar coordinates is:

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3

Find the area inside the cardioid r = a(1 + cos ) but outside the circle r = a.

Solution

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4

Find the area of the inner loop of the limacon r = a(1 + 2 cos ).

Solution

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5

Find the area enclosed by four-leaved rose r = a cos 2.

Solution



0°

15°

30°

45°

60°

75°

90°

r

a

0.87a

0.5a

0

-0.5a

-0.87a

-a

Since cos (-2) = cos 2, the equation remains unchanged when  is replaced by -, the curve is symmetric with respect to the x-axis. The equation remains unchanged when Î¸ is replaced by (180° - ), since cos 2( - ) = cos 2. Therefore, the graph is symmetric with respect to the yaxis. Because of symmetry, we can sketch the curve without recourse to point-by-point plotting.

answer


6

What is the area within the curve r 2 = 16 cos ?

Solution

 r

0° ±4

±30° ±3.72

±60° ±2.83

±90° 0

> 90° imaginary

The values in the table show that the graph is symmetrical to the origin and  ranges from -90° to 90°.

answer

V olumes of Solids of Revolution |

Applications of Integration

The solid generated by rotating a plane area about an axis in its plane is called a solid of revolution . The volume of a solid of revolution may be found by the following procedures:

Circular Disk Method

The strip that will revolve is perpendicular to the axis of revolution. In this method, the axis of rotation may or may not be part of the boundary of the plane area that is being revolved.


The disk as shown in the figure has an outer radius of xR , a hole of radius xL, and thickness dy. The differential volume is therefore  xR 2 dy -  xL2 dy and the total volume is...

The integration involved is in variable y since the derivative is dy, xR and xL therefore must be expressed in terms of y. If the axis of revolution is part of the boundary of the plane area that is being revolved, xL = 0, and the equation reduces to...


From the figure shown below, the volume can be found by the formula...

If yL = 0, we have

$

Where yU and yL are functions of x.

Cylindrical Shell Method

The strip that will revolve is parallel to the axis of revolution. The volume of revolution is obtained by taking the limit of the sum of cylindrical shell elements, each of which is equal in volume to the mean circumference times the height times the thickness.



Example 1 | V olumes of Solids of Revolution Example

1

Find the volume of the solid generated when the area bounded by the curve y2 = x, the x-axis and the line x = 2 is revolved about the x-axis.

Solution Circular Disk Method

answer

Cylindrical Shell Method

answer

Example 2 | V olumes of Solids of Revolution Example

2

Find the volume generated when the area in Example 1 will revolve about the y-axis.

Indefinite Integrals

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