Antidifferentiation: The Indefinite Intergral Chapter Five

§5.1 Antidifferetiation

§5.1 General Antiderivative of a Function

§5.1 Rules for Integrating Common Function The Constant Rule

§5.1 Rules for Integrating Common Function

Example: Solution:

§5.1 Applied Initial Value Problems An initial Value problems is a problem that involves solving a differential equation subject to a specified initial condition. For instance, we were required to find y=f(x) so that A Differential equation is an equation that involves differentials or derivatives. We solved this initial problem by finding the antiderivative And using the initial condition to evaluate C.

The population p(t) of a bacterial colony t hours after observation begins is found to be change at the rate If the population was 2000,000 bacteria when observations began, what will be population 12 hours later? Example: Solution:

§5.2 Integration by Substitution How to do the following integral?

§5.2 Integration by Substitution Think of u=u(x) as a change of variable whose differential is Then

Example: Solution: Find

Example: Solution:

Example: Solution: To be continued

Example: Solution:

Example: Solution:

§5.3 The Definite Integral and the Fundamental Theorem of Calculus

All rectangles have same width. n subintervals: Subinterval width Formula for x i :

Choice of n evaluation points

Right-endpoint approximation left-endpoint approximation

Midpoint Approximation

Example: =0.285 To be continued

= =0.385

Example: left-endpoint approximation

Midpoint Approximation Right-endpoint approximation

S 200 = S 400 = Area Under a Curve Let f(x) be continuous and satisfy f(x)≥0 on the interval a≤x≤b. Then the region under the curve y=f(x) over the interval a≤x≤b has area Where x j is the point chosen from the jth subinterval if the Interval a≤x≤b is divided into n equal parts, each of length

§5.3 The Definite Integral Riemann sum Let f(x) be a function that is continuous on the interval a≤x≤b. Subdivide the interval a≤x≤b into n equal parts, each of width,and choose a number x k from the kth subinterval for k=1, 2, …,. Form the sum Called a Riemann sum. Note: f(x)≥0 is not required

§5.3 The Definite Integral The Definite Integral the definite integral of f on the interval a≤x≤b, denoted by, is the limit of the Riemann sum as n→+∞; that is The function f(x) is called the integrand, and the numbers a and b are called the lower and upper limits of integration, respectively. The process of finding a definite integral is called definite integration. Note: if f(x) is continuous on a≤x≤b, the limit used to define integral exist and is same regardless of how the subinterval representatives x k are chosen.

§5.3 Area as Definite Integral If f(x) is continuous and f(x)≥0 for all x in [a,b],then and equals the area of the region bounded by the graph f and the x-axis between x=a and x=b If f(x) is continuous and f(x)≤0 for all x in [a,b],then And equals the area of the region bounded by the graph f and the x-axis between x=a and x=b

§5.3 Area as Definite Integral equals the difference between the area under the graph of f above the x-axis and the area above the graph of f below the x-axis between x=a and x=b This is the net area of the region bounded by the graph of f and the x-axis between x=a and x=b

§5.3 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus If the function f(x) is continuous on the interval a≤x≤b, then Where F(x) is any antiderivative of f(x) on a≤x≤b Another notation:

§5.3 The Fundamental Theorem of Calculus (Area justification ) In the case of f(X)≥0, represents the area the curve y=f(x) over the interval [a,b]. For fixed x between a and b let A(x) denote the area under y=f(x) over the interval [a,x].

By the definition of the derivative,

Differentiation Indefinite Integration Definite integration Example

§5.3 Integration Rule Subdivision Rule

§5.3 Subdivision Rule

Example Solution:

Example Solution: To be continued

§5.3 Substituting in a definite integral

2.

§5.3 Substituting in a definite integral

Example Solution:

Example Solution: