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set12 - Lecture 33 The denite integral and its...

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Lecture 33 The definite integral and its applications (cont’d) Using definite integrals instead of indefinite integrals (antiderivatives) in solving problems There is no relevant section in the textbook by Stewart for the material presented in today’s lecture. In the previous lecture, we discussed an important application of the Fundamental Theorem of Calculus II, namely, that integraldisplay b a F ( x ) dx = F ( b ) F ( a ) . (1) This is true for any differentiable function F ( x ) since, by definition, F ( x ) is an antiderivative of F ( x ). The quantity on the left hand side of the equation represents an integration of the rate of change of F ( x ), namely, F ( x ), from x = a to x = b . The quantity on the right hand side represents the net change of F ( x ) from x = a to x = b . Section 5.4 of the textbook by Stewart (Sixth Edition) discusses a number of important applica- tions for this “Net Change” property of definite integrals. And, indeed, in the previous lecture, we examined an important application to classical mechanics: Suppose that v ( t ) represents the velocity of a particle travelling on a straight line (represented by the x -axis). The relation of the position x ( t ) and the velocity function v ( t ) is, of course, v ( t ) = dx dt = x ( t ) . (2) From the Net Change property, which is simply a consequence of FTC II, we have that integraldisplay b a v ( t ) dt = integraldisplay b a x ( t ) dt = x ( b ) x ( a ) . (3) In other words, the definite integral of v ( t ) over the time interval [ a, b ] is the net displacement of the particle over this time interval. We now consider a slightly different form of the above Net Change property. We’ll integrate the velocity function from a starting time, say t = 0 to a future time t > 0. But we’ll keep this upper limit as a variable. With regard to the previous equation, we’ll replace a with 0 and b with t . This 219
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means that we’ll have to change the integration variable, t , to something else, say, s . The result is integraldisplay t 0 v ( s ) ds = integraldisplay t 0 x ( s ) ds = x ( t ) x (0) . (4) We can rewrite the above equation as follows, x ( t ) = x (0) + integraldisplay t 0 v ( s ) ds. (5) We can check this result by differentiating. The derivative of the LHS is simply x ( t ). Since x (0) is a constant, the derivative of the RHS is, by the FTC I, d dt integraldisplay t 0 v ( s ) ds = v ( t ) = x ( t ) . (6) We’ll use this idea repeatedly in this lecture. A return to the “near-earth” falling body problem We consider the motion of an object of mass m near the surface of the earth, where the force acting on it is assumed to be F ( x ) = f ( x ) i , where f ( x ) = mg. (7) As before, the positive x -axis points upward, with x = 0 denoting the surface of the earth. If we let x ( t ) denote the position of the mass at time t , then Newton’s equation of motion, F = m a becomes f = ma, (8) where a ( t ) = v ( t ) , (9) is the acceleration. Of course, in this simple case, f = mg , so that Newton’s equation becomes mg = ma a = g. (10) In other words, the acceleration of the object is constant, as we know very well.
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