# Dx lim t a z b t f x dx the same can be done when f x

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Chapter 3 / Exercise 40
College Algebra with Applications for Business and Life Sciences
Larson
Expert Verified
dx = lim t a - Z b t f ( x ) dx. The same can be done when f ( x ) is defined on [ a, b ), or even if, say f ( x ) is only defined on [ a, c ) ( c, b ] (with a hole in the middle of the interval), then Z b a f ( x ) dx = lim t c - Z t a f ( x ) dx + lim t c + Z b t f ( x ) dx Note that the above integral will only converge if both limits exist. Example 25. Proposition 1. [Comparison test for improper integrals] Suppose that f ( x ) g ( x ) on [ a, ) . If R a f ( x ) dx converges, then R a g ( x ) dx also converges. Furthermore, if R a g ( x ) dx diverges, then so does R a f ( x ) dx . Example 26. Does R 0 e - x 2 dx converge? 22
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Chapter 3 / Exercise 40
College Algebra with Applications for Business and Life Sciences
Larson
Expert Verified
Soloution: We have x 2 x on [1 , ] so e x 2 e x on [1 , ) or 1 e x 2 1 e x . So Z 0 e - x 2 dx = Z 1 0 e - x 2 dx + Z 1 e - x 2 dx Z 1 0 e - x 2 dx + Z 1 e - x dx = Z 1 0 e - x 2 dx + lim t →∞ Z t 1 e - x dx = Z 1 0 e - x 2 dx + lim t →∞ - e - t - ( - e - 1 ) = Z 1 0 e - x 2 dx + e < so our integral converges by comparison test. 17. Differential Equations A differential equation (DE) is an equation which contains a derivative expression. The order of a DE is the highest derivative which appears in the equation. Example 27. xy 00 ( x ) = sin xe y ( x ) is order 2. Another, simpler, DE is y 0 ( x ) = sin x + e x . This is easy to solve (integrate both sides), y ( x ) = - cos x + e x + C is a solution. Lets see a more exciting example. Example 28. Let P ( t ) be a population function (with t for time). After studying the population over time, and discovering a proportionality amongst the P ( t ) and its derivative. That is to say P 0 ( t ) = kP ( t ) , for some constant k . The goal now, is to obtain a formula for P ( t ) . Rearranging slightly is P 0 ( t ) P ( t ) = k Z P 0 ( t ) P ( t ) dt = Z kdt ln | P ( t ) | = kt + C e ln P ( t ) = e kt + C P ( t ) = e kt e C P ( t ) = C 0 e kt for some positive constant C that is determined by specified initial conditions (such as P (0) = 100 ). Example 29. Verify that y = x 2 ln x is a solution to x 2 y 00 - 3 xy 0 + 4 y = 0 in the region x > 0 . Solution: With y = x 2 ln x , we compute y 0 = 2 x ln x + x and y 00 = 2 ln x + 2 + 1 . Substituting into the DE, we find x 2 y 00 - 3 xy 0 + 4 y = x 2 (2 ln x + 3) - 3 x (2 x ln x + x ) + 4 x 2 ln x = 2 x 2 ln x + 3 x 3 - 6 x 2 ln x - 3 x 2 + 4 x 2 ln x = 0 A particular solution to a DE is one without arbitrary constants. That is, any particular function satisfying the DE. The solution y = x 2 ln x from the previous is example is a particular solution. Example 30. Suppose we want a solution to y 0 + 2 y = e - x with y (0) = 3 [This is called an initial value problem ]. Check that y = e - x + Ce - 2 x is a general solution. Indeed, y 0 = - e - x - 2 Ce - 2 x and - e - x - 2 Ce - 2 x + 2( e - x + Ce - 2 x = 0 . 23
Now since y (0) = 3 , we can solve for the constant C as 3 = e 0 + Ce 0 = 1 + C C = 2 so that y = e - x + 2 e - 2 x is a particular solution to the initial value problem (IVP). 17.1. Separable Differential Equations. If a DE can be expressed in the form y 0 = g ( x ) f ( y ), we say it is separable . Equivalently 1 f ( y ) dy dx = g ( x ) 1 f ( y ) dy = g ( x ) dx. A separable DE is solved by integrating both sides of this rearrangement.