The Reimann Integral is a formal limit definition of a definite integral

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1 MATH 136 The Reim Itegrl The Reim Itegrl is forml limit defiitio of defiite itegrl cotiuous fuctio f. The costructio is s follows: f ( x) dx for Reim Itegrl: Prtitio [, ] ito suitervls ech hvig the equl legth of i ( ) ( ) / givig poits = x 0 < x 1 < x <... < x =, where x i = +, for i = 0, 1,...,. Let c i e y poit i the itervl [ x i 1, x i ] for i = 1,...,. f ( x) x c 0 1 x c 1 x c 3 x c 3 4 x 4 f (c ) f (c 3 ) f (c 4 ) f (c 1 ) x c 0 1 x c 1 x c 3 x c 3 4 x 4 The, f ( x) dx f (c i )(x i x i 1 ) = f (c i ), d f ( x) dx = lim f (c i ). Wheever f is piecewise cotiuous or oly hs coutle umer of removle discotiuities, the the limit of the pproximtios exists s the umer of sudivisios icreses to ifiity. The pproximtios my or my ot coverge if there is ifiite discotiuity.

2 There re vrious wys to pick the c i. We could pick c i t rdom from [ x i 1, x i ], or we could use the right edpoit y lettig c i = x i, or we could eve use the midpoit y lettig c i = ( x i 1 + x i ) /. We ofte write (x i x i 1 ) = Δx i for chge i x, eve if they re ot eqully spced prt. The f ( x) dx f (c i ) Δx i, d f ( x) dx = lim f (c i ) Δx i i=1 Δx 0 Here, Δx 0 mes tht the legth of ll suitervls must e decresig to 0. The limit turs to, turs f (c i ) to f ( x), d turs Δx i to dx. Exmple. Use the forml limit defiitio of Reim Itegrl, with right edpoit sums, to derive formul for (mx + g) dx. Solutio. We first prtitio [, ] ito suitervls of equl legth ( ) / y i ( ) lettig = x 0 < x 1 < x <... < x =, where x i = +, for i = 0, 1,...,. We the let c i = x i for i = 1,...,. To evlute the Reim Sum pproximtio i this cse, we ( +1) shll eed the formul for the sum of the first turl umers: i =. For 10 exmple, = i = 10(11) = 55. i=1 i ( ) Now lettig f ( x) = mx + g d c i = +, we first hve (mx + g) dx = = m + m + = ( ) m + f (c i ) i m( ) m( ) = + g m ( )( + 1) = m + m + i ( ) i + g = m + + g. + g i m( ) + g m( ) ( +1) + g The tkig the limit, we oti

3 m( )( +1) (mx + g) dx = lim ( ) m + = ( ) m + = ( ) = m( ) m + m m( ) + g( ). + g + g = ( ) m + m ( ) + g = m ( )( + ) + g( ) + g (We c verify the result quickly with the Fudmetl Theorem of Clculus.) We c see tht the process is difficult eve with sic lier fuctio. Oviously, it would e prolemtic to do geerl defiite itegrls i this mer. Usig the FTC is clerly preferle. But the FTC is ctully derived usig this process. We prove this result ext. Lemm. Let F(x ) e differetile t poit. The F is cotiuous t poit. Proof. We must show tht lim F(x ) = F(), which is equivlet to showig tht x 0 = lim F(x ) F() = lim ( F (x ) F ()). But we kow tht F () exists; thus, x x lim x ( F( x) F() ) = lim x F(x ) F() x (x ) F(x ) F() = lim lim (x ) x x x = F () 0 = 0. Hece, lim x F(x ) = F() d F is cotiuous t poit. Theorem. (The Fudmetl Theorem of Clculus) Let f e cotiuous over [, ]. Let F(x ) e tiderivtive of f ( x) so tht F ( x) = f ( x) for ll x [, ]. The f (x ) dx = F() F ().

4 Proof. We prtitio [, ] ito suitervls ech hvig the equl legth of ( ) / givig poits = x 0 < x 1 < x <... < x =. Becuse F is differetile o [, ], we kow y the precedig lemm tht F is cotiuous o [, ]. We therefore c pply the Me Vlue Theorem to F o ech suitervl [ x i 1, x i ]. I other words, for i = 1,...,, there exist poits c i [ x i 1, x i ] such tht F (c i ) = F(x i) F (x i 1 ) x i x i 1 But F (c i ) = f (c i ) for i = 1,...,. Usig these c i, we the hve for ll. Thus, f ( x) dx f (c i )(x i x i 1 ) = F (c i )(x i x i 1 ) F (x = i ) F( x i 1 ) ( x i x i 1 ) = ( F( x i ) F( x i 1 )) x i x i 1 = ( F (x 1 ) F(x 0 )) + ( F(x ) F(x 1 )) ( F( x ) F( x 1 )) = F( x ) F( x 0 ) = F() F(), f ( x) dx = lim f (c i )(x i x i 1 ) = lim ( F() F() ) = F() F(). So ow it is much esier to use the FTC to evlute defiite itegrls. For exmple, (mx + g) dx = m x + g x = m + g + g m = m( ) + g( ). However the forml Reim Itegrl limit defiitio is still used to derive other itegrl formuls. We coclude with oe such derivtio.

5 Volume of Rottiol Solid: The Disc Method Let f e cotiuous with f ( x) 0 over [, ], d let R e the regio etwee the grph of f d the x -xis over [, ]. If we rotte R roud the x -xis, the we oti three-dimesiol solid. The volume of this solid is give y Vol = π ( f (x )) dx uits 3. f ( x) R Rotte R roud x -xis x Vol = π ( f ( x)) dx Proof. We prtitio [, ] ito suitervls ech hvig the equl legth of ( ) / givig poits = x 0 < x 1 < x <... < x =. Let c i [ x i 1, x i ] for i = 1,...,. The the solid is split ito slices. The volume of ech slice is pproximted y the volume of (sidewys) cylider. f (c i ) x xis The height of the i th cylider is h i = (x i x i 1 ) d the rdius is r i = f (c i ). So the volume of the i th cylider is π(r i ) h i = π( f (c i )) (x i x i 1 ). Thus, the volume of the rottiol solid is pproximted y x i 1 x i c i Vol π( f (c i )) (x i x i 1 ) = π ( ) ( f (c i )), i=1 which is Reim sum pplied to the cotiuous fuctio ( f (x )). Tkig the limit, we oti ( ) Vol = π lim ( f (c i )) = π ( f ( x) ) dx. Chllege Exercise Use the forml limit defiitio of Reim Itegrl, with right edpoit sums, to derive formul for x dx. (You my eed the formul i ( + 1)( +1) =.) 6