Integral Calculus By Jorge Samayoa §1. (Short)History The history of calculus stares several hundreds of years ago, when the mathematician Archimedes (225 B.C.) made a very important contribution. It said that the area of a 4 2 segment of a parabola is the area of a triangle with the same base and vertex and 3 3 the area of the circumscribed parallelogram. In his work, Archimedes constructed an infinite series of triangles of known area to approximate the area of such segment. Archimedes, therefore, introduces the concept of putting together different pieces of known area to get a different area. This seems to be a very basic thought, but it was very influential to the basis of calculus. Several years after Archimedes work, and several contributions of different people, there were two people that took, pretty much, the whole credit of what is know as calculus. These people were, Sir Isaac Newton and Gottfried Leibniz. Even thought the first person who gives a precise idea of calculus was Newton, the one that publish first his work know as calculus was Leibniz. At that time, there was a controversy of who invented calculus first and which country deserved the credit. A careful examination of the papers of Leibniz and Newton shows that they arrived at their results independently, with Leibniz starting first with integration and Newton with differentiation. Today, both Newton and Leibniz are given credit for developing calculus independently. It is Leibniz, however, who gave the new discipline its name. Newton called his calculus "the science of fluxions". Nowadays, independently of whose ideas we should follow, calculus is divided in two huge problems: The first problem of calculus, which is the estimation of the slope of a curve in a given point, and the second problem of calculus which is the estimation of the area under a curve in a given interval. In this document, however, we will focus in the second problem of calculus since we already went over the first problems in the last tutorial (The Geometric Approach of the Derivative). §2. The Second Problem of Calculus (The Area Problem) (http://calclab.math.tamu.edu/~jsamayoa/final_section_2 ) The second problem of calculus is finding the area of the region R that lies under the curve y = fx() from a to b (Figure 1). Figure 1: Area under the curve In order to find the area under a curve, as described in the section below, we will approximate the area by using another figure of known area; the rectangle (Figure 2). Figure 2: Area of the Rectangle One of the persons who introduced this was Jean-Gaston Darboux. However, this approach is equivalent to the approach given by Friedrich Bernhard Riemann. There are two different ways of doing this approximation. One is the Darboux upper sums, figure 3, and the Darboux lower sums, figure 4. As you might notice, the difference relies on the selection of the lengths of the approximating rectangles. Let fab:[ , ]→ be a bounded function, and let Px= (0 ,… , xn ) be a partition of [ ab , ]. Let Mfxi = sup ( ). xx∈[,]ii−1 x The upper Darboux sum of f with respect of P is n U[fP , ]=Δ∑ Mii x . i=1 where Δ=xiii(xx −−1 ). Figure 3: Upper Darboux sum. Let f :[ab , ]→ be a bounded function, and let Px= (0 ,… , xn ) be a partition of [ ab , ]. Let mfxi = inf ( ). xx∈[,]ii−1 x The lower Darboux sum of f with respect of P is n LfP[,]=Δ∑ mii x . i=1 where Δ=xxxiii( −−1 ). Figure 4: Lower Darboux sum. Theorem 1: Let AyfxabP be the area of the region R that lies under the curve = ( ) from to , and let be a Partition of [ab , ]. Then, LfP[ , ]≤≤ A UfP [ , ] Proof: (We let the proof of this result to the reader). In order to improve our approximation of the area A. Let us make a refinement of P0 in the Darboux upper sums, say P1. Notice that the difference between the approximation A using P0 and using P1 is the length of the interval[,xiix −1 ]. If we refine Pi by Pj (i<j) the approximation will be better and hxx= ii−=Δ−1 x i is getting smaller while n is getting bigger (figure 5). Figure 5: As hn→→∞0 as . This led us to the following definitions: Definition 1: The area A of the region R that lies under the graph of f ∈Cab[,] is the limit of the Darboux sums : AUfPLfP= lim [ , ]= lim [ , ] n→∞ n→∞ where Px= (,0 … , xn )is a partition of [ab , ]. Equivalently, we state de following definition. Definition 2: th Let Ai be the area of the i approximating rectangle. Then, the area A of the region R that lies under the graph of f ∈Cab[ , ] is A==lim Ainn lim[ fx (11 ) Δ+Δ++Δ x fx ( 2 ) x 2 fx ( ) x ] nn→∞ →∞ for x j ∈[,]ab. Example 1: Estimate the area under the curve y = x2 for x∈[0,1] by 1. 4 approximating rectangles by 1.1. Darboux upper sums. 1.2. Darboux lower sums. 2. n approximating rectangles, then, let n →∞. SOLUTION: 1.1 ba− 1 Δ=x = n 4 ⎛⎞113 ⎛⎞ ⎛⎞ Af≈ ⎜⎟Δ+ xf ⎜⎟ Δ+ xf ⎜⎟ Δ+ xf()1 Δ x ⎝⎠424 ⎝⎠ ⎝⎠ 1119⎛⎞ ≈+++=⎜⎟1 0.4688 4164⎝⎠ 16 1.2 ba− 1 Δ=x = n 4 ⎛⎞113 ⎛⎞ ⎛⎞ Af≈ ()0 Δ+ xf⎜⎟ Δ+ xf ⎜⎟ Δ+ xf ⎜⎟ Δ x ⎝⎠424 ⎝⎠ ⎝⎠ 1119⎛⎞ ≈+++=⎜⎟0 0.2188 416416⎝⎠ 2 ba− 1 Δ=x = nn ⎛⎞12 ⎛⎞ Af≈Δ+Δ++Δ⎜⎟ xf ⎜⎟ x f()1 x ⎝⎠nn ⎝⎠ 22 2 11⎛⎞⎛⎞ ⎛⎞ 2 ⎛⎞n ≈+++ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟ nn⎝⎠⎝⎠ ⎝⎠ n ⎝⎠ n 11(1)(21)22 2 ⎛ nn++ n ⎞ ≈+++=22()12 n ⎜ nn⎝ 6 ⎠ ⎛⎞(1)(21)1nn++ ∴ A ==lim⎜⎟ . n→∞ ⎝⎠63n §3. Definite Integral. (http://calclab.math.tamu.edu/~jsamayoa/final_section_3 ) Definition 3: Suppose f ∈Cab[,].Let us divide the interval [,]ab in n subintervals of width ba− Δ=x . Let x ==ax,,, x… , x b, be the endpoints and x***,,,,xx… x *the sample n 012n 01 2 n * points in such intervals, so xiii lies in the ixxth subinterval [−1 , ]. Then the definite integral of f from a to b is b n * f ()xdx= lim f xi Δ x ∫ n→∞ ∑ () a i=1 Since we supposed that f is continuous in [a , b], definition 3 can be states as follows. For every numberε > 0 , there is a positive integer number N such that, b n fxdx() − f x* Δ< x ε ∫ ∑ ()i a i=1 * For every n > N and for every xiiiin [xx−1 , ]. Example 2: 1 Evaluate ∫ xdx. 0 SOLUTION: 1 n * By definition f (xdx )=Δ lim f xi x . ∫ n→∞ ∑ () 0 i=1 ba− 1 Now, the width is Δ=x = . nn thus, 1 ⎛⎞12 ⎛⎞ xdx≈Δ+Δ++Δ f⎜⎟ x f ⎜⎟ x f()1 x ∫ ⎝⎠nn ⎝⎠ 0 11⎛⎞⎛⎞⎛⎞ 2 ⎛⎞n ≈+++⎜⎟⎜⎟⎜⎟ ⎜⎟ nn⎝⎠⎝⎠⎝⎠ n ⎝⎠ n 11(1)⎛⎞nn+ ≈+++=22()1 2 n ⎜⎟ nn⎝⎠2 1 ⎛⎞(1)1n + ∴=xdx lim = . ∫ n→∞ ⎜⎟ 0 ⎝⎠22n §3.1. Properties of the definite integral. Due the nature of this document, we will state some of the most important properties of the definite integral, and the proofs of such properties are lived as exercise for the student. Properties of the Integral. b 1. ∫ cdx=− c( b a ), where c is any constant. a bbb 2. ∫∫∫f ()x±= gxdx () f () xdx ± gxdx () aaa bb 3. ∫∫cf( x ) dx= c f ( x ) dx , where c is any constant. aa bc c 4. ∫∫∫f ()xdx+= f () xdx f () xdx with cab∈( , ) . ab a 5. If mfxM≤≤( ) for axb ≤≤, then b mb()−≤ a∫ f ()(). xdx ≤ M b − a a Example 3: 1 Use properties of the integral to evaluate ∫ (2+ 4x )dx . 0 1111 ∫∫∫∫(2+=+=−+ 4x )dx 2 dx 4 xdx 2(1 0) 4 xdx 0000 by example 2, 1 ⎛⎞1 ∫ (2+=+= 4xdx ) 2 4⎜⎟ 4 0 ⎝⎠2 Definition 4 (Fundamental Theorem of Calculus) Supposef ∈Cab [ , ] . Then, x 1. If gx( )==∫ ftdt ( ) , then g'( x ) f ( x ). a b 2. ∫ f (xdx )=− Fb ( ) Fa ( ), where F is any antiderivative of f, that is, F' = f . a Proof: Since this is a support document for out tutorial, the proof of this will be in other document with different goal. §4. Indefinite Integrals. (http://calclab.math.tamu.edu/~jsamayoa/final_section_4 ) The goal of this section is to familiarize the student with the evaluation of indefinite integrals. Therefore, we will just present a table with several indefinite integrals so that the student might have a comprehension of the computational procedure of solving indefinite integrals. Because of the relation given by the Fundamental Theorem of Calculus between antiderivatives and integrals, the notation ∫ f (xdx ) is traditionally used for an antiderivatives of f and is called indefinite integral. Thus, ∫ f ()xdx== Fx () means F'() x f () x xdx22⎛⎞ For example, we can write ∫ xdx= ++= C, because ⎜⎟ C x . 22dx ⎝⎠ Table of Integrals. Example 4: x5 a) ()3cos3x4 +=++xdx sin x C ∫ 5 ⎛⎞2 2 b) ⎜⎟tanx +=++dx sec x 2ln x C ∫ ⎝⎠x ⎛⎞4 c) edeCθθ+=++θθ4sin−1 ∫ ⎜⎟2 ⎝⎠1−θ d) Use the Fundamental Theorem of Calculus, to 2 ⎛⎞2 evaluate 23x4 +−xdx . ∫ ⎜⎟2 0 ⎝⎠1+ x 2 2 ⎛⎞2 xx52 23xx41+− dx = 2 + 3 − 2tan− x ∫ ⎜⎟2 0 ⎝⎠152+ x 0 52 ⎛⎞22 −1 = ⎜⎟2+− 3 2 tan 2 − 0 ⎝⎠52 64 =+−6 2 tan−1 2 5 94 =−2 tan−1 2 5 §5 Area between two curves. (http://calclab.math.tamu.edu/~jsamayoa/final_section_5 ) In section 2 we defined and calculated areas of regions that lie under curves. In this section we will use integrals to find areas of region between two curves.