Presentation on theme: "In this handout, 4. 7 Antiderivatives 5"— Presentation transcript:
1 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
2 AntiderivativesDefinition: 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 fon an interval I, then the most generalantiderivative of f on I isF(x) + Cwhere C is an arbitrary constant.F is an antiderivative of f
3 Table of particular antiderivatives FunctionAntiderivativexr, r -1xr+1/(r+1)1/xln |x|sin x- cos xexcos xaxax / ln asec2 xtan xtan-1 xsin-1 xaf + bgaF + bGIn 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 .
4 Computing Antiderivatives ProblemSolutionUse here the formula (aF + bG)’ = aF’ + bG’
5 Analyzing the motion of an object using antiderivatives ProblemA particle is moving with the given data. Find the position of the particle.a(t) = t -3t2, v(0) = 2, s(0) = 5v(t) is the antiderivative of a(t): v’(t) = a(t) = t -3t2Antidifferentiation gives v(t) = 10t + 1.5t2 – t3 + Cv(0) = 2 implies that C=2; thus, v(t) = 10t + 1.5t2 – t3 + 2s(t) is the antiderivative of v(t): s’(t) = v(t) = 10t + 1.5t2 – t3 + 2Antidifferentiation gives s(t) = 5t t3 – 0.25t4 + 2t + Ds(0) = 5 implies that D=5; thus, s(t) = 5t t3 – 0.25t4 + 2t + 5Solution
6 Evaluation TheoremTheorem: If f is continuous on the interval [a, b], thenwhere F is any antiderivative of f, that is, F’=f.Example:since F(x)=1/3 * x3 is an antiderivative of f(x)=x2
7 Indefinite IntegralIndefinite integral is a traditional notation for antiderivativesNote: 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:
9 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 ofis
10 Average value of a function The average value of function f on the interval [a, b] is defined asNote: For a positive function, we can think of this definition as saying area/width = average heightExample: Find the average value of f(x)=x3 on [0,2].
11 The Mean Value Theorem for Integrals Theorem: If f is continuous on [a, b], then there exists a number c in [a, b] such thatExample: 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