Chapter 1 Topics in Analytic Geometry

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 71(a)(b)(c)∫ ba∫ ba∫ ba∫ bkr(t)dt = k r(t)dta[r 1 (t)+r 2 (t)]dt =[r 1 (t)−r 2 (t)]dt =∫ ba∫ bar 1 (t)dt+r 1 (t)dt−∫ ba∫ bar 2 (t)dtr 2 (t)dtAntiderivatives of Vector-Valued FunctionsAn antiderivative for a vector-valued function r(t) is a vector-valued function R(t) suchthatR ′ (t) = r(t) (4.17)We express Equation(4.17) using integral notation as∫r(t)dt = R(t)+C (4.18)where C represents an arbitrary constant vector. Note that the vector R(t) + C is alsocalled the indefinite integral of r(t).Since differentiation of vector-valued functions can be performed componentwise, it followsthat antidifferentiation can be done this way as well.∫Example 4.13 Evaluate the indefinite integral (2ti+3t 2 j+sin2tk)dt.Solution .........Mostofthefamiliarintegrationpropertieshavevector counterparts. Forexample, vectordifferentiation and integration are inverse operations in the sense that[∫ddt]r(t)dt = r(t) and∫r ′ (t)dt = r(t)+C (4.19)Moreover, if R(t) is an antiderivative of r(t) on an interval containing t = a and t = b, thenwe have the following vector form of the Fundamental Theorem of Calculus:∫ ba] b ∫r(t)dt = R(t) =a] br(t)dt = R(b)−R(a) (4.20)aExample 4.14 Evaluate the definite integralSolution .........∫ 10(sinπti+(6t 2 +4t)j ) dt.Example 4.15 Find r(t) given that r ′ (t) = 〈3,2t〉 and r(1) = 〈2,5〉.Solution .........