Lecture 21: Antiderivatives. Definition and computation
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1 Lecture 21: Antiderivatives. Definition and computation Victoria LEBED, MA1S11A: Calculus with Applications for Scientists November 28, 2017
2 1 The notion of antiderivative Today we are starting the last chapter of our course: Integral calculus. We shall firstlearn whatare itsbasicobjects,and how todeal withthem. Later we shallsee whythey are usefulin science. It is often important for applications to solve the problem reverse to differentiation: find a function if we know only its derivative. Example. You see the instantaneous velocity on the speedometer all the time, and wonder how faryou progressed over the given periodof time. Mathematically,from x (t)you wanttodeduce x(t 1 ) x(t 2 ). Definition. A functionfiscalledan antiderivative of afunctionfif df(x) dx = f(x). Ifsuch anfexists, thefunctionfis calledintegrable.
3 1 The notion of antiderivative Definition. A functionfiscalledan antiderivative of afunctionfif df(x) dx = f(x). Recall a fact we have learned about derivatives: Theorem. f (x) = g (x)on (a,b) f(x) = g(x)+con (a,b) for some c R. It can be reformulated in terms of antiderivatives: Theorem 1. F 1 and F 2 are twoantiderivativesof f on (a,b) F 1 (x) = F 2 (x)+c on (a,b) for some C R. In words, theantiderivativeof a functionon aninterval, ifexists, iswell defined up to a constant. Iffis definedon a disjointunionof intervals, then separateconstants should be taken for these intervals.
4 1 The notion of antiderivative Iffis definedon a disjointunionof intervals, then separateconstants should be taken for these intervals. Example. We have computed the following derivative: (ln x ) = 1 on (,0) (0,+ ). x It means thatln x is anantiderivativeof 1 x. Butthen the function { ln x + 3 π, x < 0, F(x) = ln x e, x > 0 is anotherantiderivativeof 1 x.
5 1 The notion of antiderivative There is an alternative integral notation for saying that F is an antiderivatives of f: f(x)dx = F(x)+C. The expression f(x)dx is calledthe indefinite integral. By definition,( f(x)dx) = f(x). The constants C should behandled withcare. Forinstance, xdx = x2 +C = xdxdx = x Cx+D. Also, you should leave the arbitraryconstant tillthe very end of the computation, to avoid something like this: ( ) x x 2 2 +C = 2xdx = 2 xdx = 2 2 +C = x 2 +2C, whichyields C = 2C, soc = 0.
6 2 Computing antiderivatives In simplest situations, antiderivatives are computed by recognising derivatives, since integration and differentiation are inverse operations: differentiation formula integration formula (x r ) = rx r 1 rx r 1 dx = x r +C sin x = cosx cosxdx = sinx+c cos x = sinx sinxdx = cosx+c tan x = 1 cos 2 x cot x = 1 sin 2 x 1 x 2 arcsin x = 1 arctan x = 1 (log a x) = 1 xlna (a x ) = a x lna dx dx 1+x 2 dx cos 2 x = tanx+c dx sin 2 x = cotx+c dx = arcsinx+c 1 x 2 1+x 2 = arctanx+c xlna = log a x+c a x lnadx = a x +C For this reason, it is important to learn the derivatives of basic functions.
7 2 Computing antiderivatives For not-in-the-differentiation-table functions, integration is far from obvious. What is,say,the antiderivativeof f(x) = 1 sin(x)? According to the brick-and-mortar approach, in order to learn how to integrate, it is crucial to understand relations between integration and gluing operations for functions. The following results are translations of the corresponding properties of derivatives. Theorem 2. Suppose that the functions f(x) and g(x) are integrable, and c Risaconstant. Then cf(x)dx = c f(x)dx, [f(x)+g(x)]dx = f(x)dx+ g(x)dx, [f(x) g(x)]dx = f(x)dx g(x)dx. The statements can bewrittenin amore explicitform, e.g., f(x)dx = F(x)+C = cf(x)dx = cf(x)+c.
8 3 u-substitution Further, from the chain rule for derivatives, we deduce the following important integration rule: Theorem 3(u-substitution). Suppose that F(x) is an antiderivative of f(x). Thenthe functionf(g(x))g (x)isintegrable, and [f(g(x))g (x)]dx = F(g(x))+C. Proof. Applying the chain rule, we obtain (F(g(x))) = F (g(x))g (x) = f(g(x))g (x). Example 1. Let us evaluate the integral (x 2 +1) 50 2xdx. Letg(x) = x Noticingthat(x 2 +1) = 2x,wecanwritetheintegralas (x 2 +1) 50 2xdx = g(x) 50 g (x)dx = g(x)51 +C = (x2 +1) 51 +C
9 3 u-substitution Example 2. Let us evaluate the integral cosx sin 2 x dx. Denotingg(x) = sinx, we get cosx sin 2 x dx = 1 g(x) 2g (x)dx = 1 g(x) +C = 1 sinx +C. Always verifyyour computationsby differentiation: ( 1 ) = ( 1 sinx sin 2 x ) sin x = 1 cosx sin 2 cosx = x sin 2 x. Example 3. Let us evaluate the integral Denotingg(x) = x, we get cos x x dx. cos x dx = cosg(x) 2g (x)dx = 2sing(x)+C = 2sin x+c. x
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