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Chapter 6: Calculus~ Hughes-Hallett. Constructing the Antiderivative Solving (Simple) Differential Equations The Fundamental Theorem of Calculus (Part 2). The Derivative is --. Physically- an instantaneous rate of change.
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Chapter 6: Calculus~Hughes-Hallett • Constructing the Antiderivative • Solving (Simple) Differential Equations • The Fundamental Theorem of Calculus (Part 2)
The Derivative is -- • Physically- an instantaneous rate of change. • Geometrically- the slopeof thetangent lineto thegraph 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 = cf(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.
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(x1) = y1 substi-tute the values into the function and solve for c: y = F(x) + C y1 = F(x1) + C C = y1 - F(x1)
Example: Solve, dr/dp = 3 sin pwith 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) = ax f(t)dt is an antiderivative of f, and equivalently:
