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**In this handout, 4. 7 Antiderivatives 5**

In this handout, 4.7 Antiderivatives 5.3 Evaluating Definite Integrals 5.4 The Fundamental Theorem of Calculus

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Antiderivatives Definition: A function F is called an antiderivative of f on an interval I if F’(x) = f(x) for all x in I. Example: Let f(x)=x3. If F(x) =1/4 * x4 then F’(x) = f(x) Theorem: If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F(x) + C where C is an arbitrary constant. F is an antiderivative of f

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**Table of particular antiderivatives**

Function Antiderivative xr, r -1 xr+1/(r+1) 1/x ln |x| sin x - cos x ex cos x ax ax / ln a sec2 x tan x tan-1 x sin-1 x af + bg aF + bG In the last and highlighted formula above we assume that F’ = f and G’ = g. The coefficients a and b are numbers. These rules give particular antiderivatives of the listed functions. A general antiderivative can be obtained by adding a constant .

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**Computing Antiderivatives**

Problem Solution Use here the formula (aF + bG)’ = aF’ + bG’

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**Analyzing the motion of an object using antiderivatives**

Problem A particle is moving with the given data. Find the position of the particle. a(t) = t -3t2, v(0) = 2, s(0) = 5 v(t) is the antiderivative of a(t): v’(t) = a(t) = t -3t2 Antidifferentiation gives v(t) = 10t + 1.5t2 – t3 + C v(0) = 2 implies that C=2; thus, v(t) = 10t + 1.5t2 – t3 + 2 s(t) is the antiderivative of v(t): s’(t) = v(t) = 10t + 1.5t2 – t3 + 2 Antidifferentiation gives s(t) = 5t t3 – 0.25t4 + 2t + D s(0) = 5 implies that D=5; thus, s(t) = 5t t3 – 0.25t4 + 2t + 5 Solution

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Evaluation Theorem Theorem: If f is continuous on the interval [a, b], then where F is any antiderivative of f, that is, F’=f. Example: since F(x)=1/3 * x3 is an antiderivative of f(x)=x2

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Indefinite Integral Indefinite integral is a traditional notation for antiderivatives Note: Distinguish carefully between definite and indefinite integrals. A definite integral is a number, whereas an indefinite integral is a function (or family of functions). The connection between them is given by the Evaluation Theorem:

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**Some indefinite integrals**

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**The Fundamental Theorem of Calculus**

The theorem establishes a connection between the two branches of calculus: differential calculus and integral calculus: Theorem: Suppose f is continuous on [a, b]. Example of Part 1: The derivative of is

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**Average value of a function**

The average value of function f on the interval [a, b] is defined as Note: For a positive function, we can think of this definition as saying area/width = average height Example: Find the average value of f(x)=x3 on [0,2].

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**The Mean Value Theorem for Integrals**

Theorem: If f is continuous on [a, b], then there exists a number c in [a, b] such that Example: Find c such that fave=f(c) for f(x)=x3 on [0,2]. From previous slide, f(c)=fave=2. Thus, c3=2, so

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