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Unformatted text preview: Chapter 3. Integration 3.1 Indefinite Integral Integration can be considered as the antithesis of dif ferentiation, and they are subtly linked by the Fun damental Theorem of Calculus . We first in troduce indefinite integration as an “inverse” of dif ferentiation. 3.1.1 Antiderivatives A (differentiable) function F ( x ) is an antiderivative of a function f ( x ) if F ( x ) = f ( x ) for all x in the domain of f. 2 MA1505 Chapter 3. Integration The set of all antiderivatives of f is the indefinite integral of f with respect to x , de noted by Z f ( x ) dx. Terminology: f : integrand of the integral x : variable of integra tion 3.1.2 Constant of Integration Any constant function has zero derivative. Hence the antiderivatives of the zero function are all the constant functions. If F ( x ) = f ( x ) = G ( x ), then G ( x ) = F ( x ) + C , 2 3 MA1505 Chapter 3. Integration where C is some constant. So Z f ( x ) dx = F ( x ) + C. C here is called the constant of integration or an arbitrary constant . Thus, Z f ( x ) dx = F ( x ) + C means the same as d dx F ( x ) = f ( x ) . In words, indefinite integral and antiderivative (of a func tion) differ by an arbitrary constant. 3 4 MA1505 Chapter 3. Integration 3.1.3 Integral formulas 1. Z x n dx = x n +1 n + 1 + C, n 6 = 1 , n rational Z 1 dx = Z dx = x + C (Special case, n = 0) 2. Z sin kxdx = cos kx k + C 3. Z cos kxdx = sin kx k + C 4. Z sec 2 xdx = tan x + C 5. Z csc 2 xdx = cot x + C 6. Z sec x tan xdx = sec x + C 7. Z csc x cot xdx = csc x + C 4 5 MA1505 Chapter 3. Integration 3.1.4 Rules for indefinite integration 1. Z kf ( x ) dx = k Z f ( x ) dx , k = constant (independent of x ) 2. Z f ( x ) dx = Z f ( x ) dx (Rule 1 with k = 1) 3. Z £ f ( x ) ± g ( x ) / dx = Z f ( x ) dx ± Z g ( x ) dx 3.1.5 Example Find the curve in the xyplane which passes through the point (9 , 4) and whose slope at each point ( x,y ) is 3 √ x . Solution. The curve is given by y = y ( x ), satisfying (i) dy dx = 3 √ x and (ii) y (9) = 4. 5 6 MA1505 Chapter 3. Integration Solving (i), we get y = Z 3 √ xdx = 3 x 3 / 2 3 / 2 + C = 2 x 3 / 2 + C. By (ii), 4 = (2)9 3 / 2 + C = (2)27 + C, C = 4 54 = 50. Hence y = 2 x 3 / 2 50. 6 7 MA1505 Chapter 3. Integration 3.2 Riemann Integrals 3.2.1 Area under a curve Let f = f ( x ) be a nonnegative continuous function f = f ( x ) on an interval [ a,b ]. Partition [ a,b ] into n consecutive subintervals [ x i 1 ,x i ] ( i = 1 , 2 ,...,n ) each of length Δ x = b a n , where we set a = x , b = x n , and x 1 ,x 2 , ··· ,x n 1 to be successive points between a and b with x k x k 1 = Δ x ....
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