The Definite Integral

Transcription

1 The Defiite Itegrl A Riem sum R S (f) is pproximtio to the re uder fuctio f. The true re uder the fuctio is obtied by tkig the it of better d better pproximtios to the re uder f. Here is the forml defiitio, give i our text, of the defiite itegrl. Defiitio. Let f be fuctio defied o the itervl [, b] where<b.thefis Riem itegrble o [, b] ifthereisumberl stisfyig the followig: For every ">0, there is >0 such tht for ll smpled prtitios (, S), if <,the f(s i ) x i L <". Recllig tht R S (f) f(s i) x i,wewrite The it L is deoted by!0 RS (f)!0 f(s i ) x i L. f(x) dx d is clled the defiite itegrl of f o [, b]. Z Remrk. The symbol is clled the itegrl sig. I the ottio f(x) dx, the fuctio f(x) is clled the itegrd. The rel umbers d b re clled the its of itegrtio; is clled the lower it d b is clled the upper it. Theorem. If f is cotiuous o [, b], the f is Riem itegrble o [, b]; tht is, the defiite itegrl R b f(x) dx exists. Moreover, if f hs t most, fiite umber of discotiuities i [, b], the f is Riem itegrble o [, b]. Sice the fuctio f(x) x is cotiuous, we coclude tht R 4 x dx exists. How c we evlute R 4 x dx? The ext theorem will give us method for ctully evlutig defiite itegrls. Theorem 3. Suppose tht f is Riem itegrble o [, b]. Let (, S ) be sequece of smpled prtitios of the itervl [, b] such tht! 0. The Recll the followig results:! RS (f) Theorem 4. The followig fudmetl sums hve the give closed form solutios:. i ++ + (+). 3. i (+)(+) 6 h i i (+) roblem 5. Fid closed form solutio for i(i + ).

2 rocedure for Evlutig the Defiite Itegrl R b f(x) dx usig Theorem 3: Step : Let be the umber of rectgles. The width of ech of these rectgles is x i x b. Step : Let be the prtitio {x 0,x,...x x i + i x + i b,x } stisfyig for i 0,,...,. Thus, x x, x + x, x + x,..., x + x. Step 3: ick smple poit s i i ech subitervl [x i,x i ]. If oe picks the left edpoits of ech subitervl, the s i x i +(i ) x. If oe picks the right edpoits of ech subitervl, the s i x i + i x. If oe picks the midpoits of ech subitervl, the s i x i +x i + i x. Step 4: Usig x obtied i Step d the smple poits s i selected i Step 3, the desired Riem sum is f(s i ) x. Step 5: Fid closed form solutio of this Riem sum f(s i ) x. Step 6: Determie the it of this Riem sum s goes to ifiity. Thus, Z ( b ) f(x) dx f(s i ) x.! roblem 6. Evlute the defiite itegrl R b f(x) dx for the cotiuous fuctio f(x) x. Solutio. We pply Theorem 3. Our itervl is [, b]. We brek our solutio ito 6 steps: Step : Let be the umber of rectgles. The width of ech of these rectgles is x i x b. Step : Thus our prtitio is geerted by the formul x i + i x + i b. Step 3: ick smple poit s i i ech subitervl [x i ech subitervl. Thus, we will use,x i ]. Let us pick the right edpoits of Step 4: Thus the desired Riem sum is s i x i + i x + pple + (b (b )i. )i (b ).

3 3 Step 5: Fid closed form solutio of this Riem sum (b ) (b ) (b ) pple + (b f(s i ) x: )i (b ) + (b " + " + )i (b ) (b )i # # i pple (b ) (b ) ( + ) + pple b + (b ) + Step 6: Determie the it of this Riem sum s goes to ifiity obtiig: Hece, xdx b pple b xdx (b ) +! pple b (b ) + (b ) (b + ) b.. Oe c lso show tht + kdx kb k for y costt k. ropositio 7. Suppose tht <b, d k is costt. The, the cotiuous fuctios f(x) x d f(x) k re Riem itegrble. Moreover, xdx b d kdx kb k. roblem 8. Evlute the defiite itegrl R 4 f(x) dx for the cotiuous fuctio f(x) x. Solutio. We pply Theorem 3. Our itervl is [, b] [, 4]. We brek our solutio ito 6 steps: Step : Let be the umber of rectgles. The width of ech of these rectgles is x i x b 4 3. Step : Thus our prtitio is geerted by the formul x i + i x +i 3.

4 4 Step 3: ick smple poit s i i ech subitervl [x i ech subitervl d thus we will use s i x i + i Step 4: Thus the desired Riem sum is,x i ]. Let us pick the right edpoits of x + 3i. Step 5: Fid closed form solutio of this Riem sum 3 + 3i! 3 +3i 3 3 ( +3i) i +9i i 3. i +9 f(s i ) x:! i 3 3 ( + ) ( + )( + ) ( + )( + ) ( + ) + ( + ) ( + )( + ) Step 6: Determie the it of this Riem sum s goes to ifiity obtiig: Z 4 f(x) dx 3 ( + ) ( + )( + )! ! Hece, Z 4 x dx. I similr mer, oe c show tht x dx 3 b Remrk. The vrible x i the defiite itegrl R b f(x) dx is dummy vrible. Thus, f(x) dx f(t) dt.

5 5 roperties of the Defiite Itegrl I the followig, we ssume tht f d g re Riem itegrble fuctios. Defiitio 9. Let f(x) be fuctio defied o the itervl [, b] where<b.thewedefie Z b f(x) dx Defiitio 0. Let f(x) be fuctio d let be rel umber. The Theorems, 3, 5 below c be prove usig Theorem 3. Z f(x) dx 0. Theorem (Lier roperty). Let f(x) d g(x) be fuctios defied o the itervl [, b], d let c d k be costts. The followig lierity property holds: roblem. Evlute (cf(x)+kg(x)) dx c Solutio. We evlute R b (cx + k) dx s follows: f(x) dx + k (cx + k) dx where c d k re costts. g(x) dx. (cx + k) dx c xdx+ kdx by lierity c b + kb k by ropositio 7. Theorem 3 (Itervl Sum roperty). Let f(x) be fuctio defied o the itervls [, c], [c, b], d [, b]. The roblem 4. Evlute Z 3 f(x) dx Z c f(x) dx + x dx. [This will be doe i clss.] c Theorem 5 (Compriso roperties). Let f(x) d g(x) be fuctios defied o [, b]. The. If 0 pple f(x) for ll x i [, b], the 0 pple. If f(x) pple g(x) for ll x i [, b], the f(x) dx pple g(x) dx. 3. If M d m re costts such tht m pple f(x) pple M for ll x i [, b], the m(b ) pple f(x) dx pple M(b ). 4. f(x) dx pple f(x) dx.