Constructing the Antiderivative Solving (Simple) Differential Equations The Fundamental Theorem of Calculus (Part 2) Chapter 6: Calculus~ Hughes-Hallett.

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Constructing the Antiderivative Solving (Simple) Differential Equations The Fundamental Theorem of Calculus (Part 2) Chapter 6: Calculus~ Hughes-Hallett

The Derivative is -- Physically- an instantaneous rate of change. Geometrically- the slope of the tangent line to the graph of the curve of the function at a point. Algebraically- the limit of the difference quotient as h  0 (if that exists!). In symbols:

Review: The Definite Integral Physically - is a summing up Geometrically - is an area under a curve Algebraically - is the limit of the sum of the rectangles as the number increases to infinity and the widths decrease to zero:

Review of The Fundamental Theorem of Calculus (Part 1) If f is continuous on the interval [a,b] and f(t) = F’(t), then: In words: the definite integral of a rate of change gives the total change.

Two Fundamental Properties of the Antiderivative If F’(x) = 0 on an interval, then F(x) = C on this interval If F(x) and G(x) are both antiderivatives of f(x) then F(x) = G(x) + C

Properties of Antiderivative: 1.  [f(x)  g(x)]dx =  f(x)dx   g(x)dx (The antiderivative of a sum is the sum of the antiderivatives.) 2.  cf(x)dx = c  f(x)dx (The antiderivative of a constant times a function is the constant times the anti- derivative of the function.)

The Definition of Differentials (given y = f(x)) 1. The Independent Differential dx: If x is the independent variable, then the change in x,  x is dx; i.e.  x = dx. 2. The Dependent Differential dy: If y is the dependent variable then: i.) dy = f ‘(x) dx, if dx  0 (dy is the derivative of the function times dx.) ii.) dy = 0, if dx = 0.

Using the differential with the antiderivative.

Solving First Order Ordinary Linear Differential Equations To solve a differential equation of the form dy/dx = f(x) write the equation in differential form: dy = f(x) dx and integrate:  dy =  f(x)dx y = F(x) + C, given F’(x) = f(x) If initial conditions are given y(x 1 ) = y 1 substi- tute the values into the function and solve for c: y = F(x) + C  y 1 = F(x 1 ) + C  C = y 1 - F(x 1 )

Example: Solve, dr/dp = 3 sin p with r(0)= 6, i.e. r= 6 when p = 0 Solution:

The Fundamental Theorem of Calculus (Part 2) If f is a continuous function on an interval, & if a is any number in that interval, then the function F, defined by F(x) =  a x f(t)dt is an antiderivative of f, and equivalently:

Example: