Chapter 4 | Applications of Derivatives
493
Therefore,
∫ tan x cos x = ∫ sin x =−cos x + C .
⎛ ⎝ 4 x 3 −5 x 2 + x −7 ⎞
Evaluate ∫
4.51
⎠ dx .
Initial-Value Problems We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations. A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation (4.9) dy dx = f ( x ) is a simple example of a differential equation. Solving this equation means finding a function y with a derivative f . Therefore, the solutions of Equation 4.9 are the antiderivatives of f . If F is one antiderivative of f , every function of the form y = F ( x )+ C is a solution of that differential equation. For example, the solutions of dy dx =6 x 2 are given by y = ∫ 6 x 2 dx =2 x 3 + C . Sometimes we are interested in determining whether a particular solution curve passes through a certain point ( x 0 , y 0 ) —that is, y ( x 0 ) = y 0 . The problem of finding a function y that satisfies a differential equation (4.10) dy dx = f ( x ) with the additional condition (4.11) y ( x 0 ) = y 0 is an example of an initial-value problem . The condition y ( x 0 ) = y 0 is known as an initial condition . For example, looking for a function y that satisfies the differential equation dy dx =6 x 2 and the initial condition y (1) =5 is an example of an initial-value problem. Since the solutions of the differential equation are y =2 x 3 + C , to find a function y that also satisfies the initial condition, we need to find C such that y (1) =2(1) 3 + C =5. From this equation, we see that C =3, and we conclude that y =2 x 3 +3 is the solution of this initial-value problem as shown in the following graph.
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