Math 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED


 Reynard Wilkerson
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1 Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE
2 Approimting the re under curve given the type of Riemnn sums. Strt by finding the width of ech rectngle. Recll prtition of closed intervl [ b, ] is finite subset of [ b, ] tht contins the points nd b. The lengths of these subintervls my or my not be equl. If the lengths re equl, it is clled regulr prtition nd Δ = b n.. Now find the height of the rectngles. Use the pproprite point in ech subintervl to compute the vlue of the function t ech of these points (gives the heights of the rectngles). 3. Find the re of ech rectngle nd dd them up. S* P = f Δ + f Δ f Δ ( ) ( ) ( ) ( ) n n
3 Remember tht continuous function on closed intervl [, ] vlue m i nd mimum vlue M i. Definition: Let f be continuous function on [,b ] nd P {,,...,n} [,b ]. The sum: U f ( P) = M Δ + M Δ + M 3 Δ M n Δ n is clled the upper sum of f with respect to the prtition P. The sum: ( ) 3 3 L P = m Δ + m Δ + m Δ m Δ f n n is clled the lower sum of f with respect to the prtition P. i i lwys tkes on minimum = be prtition of
4 Find the lower Riemnn Sum for f () = 4 on [,,, ].. 5 b y c. d. e. ½
5 Find the upper nd lower sums for f ( ) = , 5.,,, 5. Lower Sum = 4 3 Upper Sum = 3 3
6 Given f ()d = 3, 3 6 f ()d = 5, f ()d = 9, find 3 6 f ()d 6 f ()d f ()d 6
7 Given tht f ( ) 7 7 d = 8 nd f ( ) d = 3. Wht is f ( ) 5 Given f ( ) d =, g ( ) d = 4, f ( ) d = 6, f ( ) d = 8. 6 Wht is ( ) 5 f d 5 5 6
8 If the curve is sometimes negtive, then one cn split the region into pieces using the roots of the function s the limits of the integrl. Consider the function whose grph is given below: Here, Are of Ω = b f ( ) d nd Are of Ω = f( ) d = f( ) d c. Emple 4: Given the grph of f, if the re of Ω is nd re of b c b Ω is 8, find f ( ) c d.
9 Emple: Given ( ) f d = 5, f ( ) d = 4, ( ) 5 f d = 4. 4 Find the re under the curve from =  to = 5.
10 Try this: Try these: A C B The re of region A is 6, region B is 7/8 nd of region C is Find f ()d. b. Wht is the re between f() nd the is between = nd =8?
11 If f is n odd function, then f ( ) d=. If f is n even function, then ( ) = ( ) f d f d.
12 if f () is n odd function, f ()d = if f () is n even function, f ()d = f ()d
13 Recll from the previous section tht: Section 6. The Fundmentl Theorem of Clculus For function f which is continuous on [,b ], there is one nd only one number tht stisfies the inequlity L ( P) I U ( P), for ll prtitions P of [,b ]. This unique number I is f f b clled the definite integrl (or just the integrl) of f from to b nd is denoted by f () d. This number cn be positive, negtive or zero. Let f be continuous function over the intervl [,b ]. Define new function by ( ) ( ) F = f t dt. Here, the upper limit vries between nd b.
14 If f hppens to be nonnegtive function, then F( ) cn be seen s the re under the grph of f from to. We cn think of F( ) s the ccumulted re function. So for emple, if ( ) ( ) F = f t dt then this is re under the function f from = to =. if ( 5) ( ) 5 F = f t dt then this is re under the function f from = to = 5.
15 Emple : If f is the function whose grph is given below, nd F ( ) = f ( t) dt, find ( 5) F. Emple : Given ( ) F = t cos ( t) dt, find F(7). 7
16 If we consider n integrl s n ccumultion of re, then the derivtive of the integrl is rte of chnge of n ccumultion of n re. Therefore, if F( ) = f ( t) dt, then F ( ) = ( ) ' f. F() = Where f is positive, F in incresing Where f is negtive, F is decresing Where f is zero, F hs possible m, min or inflection point Where f is incresing, F is concve up Where f is decresing, F is concve down A theorem connects differentil clculus nd integrl clculus.
17 Fundmentl Theorem of Clculus Prt If f is continuous function over the intervl [ ],b, then the function ( ) ( ) F = f t dt is continuous on [,b ] nd differentible on (,b ). Moreover, F' ( ) = f ( t) dt = f ( ) for ll in (,b ). Note: If = then ( ) ( ) F = f t dt =. d d, Emple 3: Find '( ) F = ( t + t) dt. F given ( )
18 Emple 4: Find the derivtive of F ( ) = dv. Then find F '(). v + 4 Emple 5: Find d 5 cos ( s) ds d. Then find F '(4 π ).
19 Emple 6: Find d sin(t) dt d. Recll the following definite integrl property from Section 6.: f() d= f() d b b Emple 7: Find '( ) F given ( ) F = 3t + dt.
20 Other times if the upper limit is not just, we ll need to use the chin rule. Tht is, d ( v( ) ) f() t dt f( v( )) v'( ) d = Emple 8: Find '( ) F given ( ) F = tsin() t dt. Try this one: Find '( ) F given F ( ) = 5 dt. + t π
21 Emple 9: Find d d 4 t+ e dt. π Try this one: sin d d tdt And yet, other times, if both limits of integrtion re functions of we ll pply the following rule: d d ( v( ) ) u( ) f() t dt = f( v( )) v'( ) f( u( )) u'( ) Emple : Given F ( ) = t dt, find ( ) 3 F'.
22 dt. d Emple : Find ( + t) d sin We stted the equtions below t the beginning of this section, so keep in mind tht: ( ) ( ) F = f t dt F is the integrl of f nd d F' ( ) = f ( t) dt = f ( ) d f is the derivtive of F Emple : Let f be continuous function stisfying 3 ( ) + = f t dt.. Stte F(). b. Find F' ( ) = f( ). c. Find ( ) F'' f '( ) =.
23 Remember: where f is positive, F is incresing where f is negtive, F is decresing where f is zero, F hs possible m, min or inflection point where f is incresing, F is concve up where f is decresing, F is concve down F = ( t 4 t ) dt. Emple 3: Let the function F be defined by ( ) 3. Find ny criticl numbers for F. b. Discuss the concvity of F.
24 Let A() = f ( ) t dt, 3 4, where f is the function grphed. 3. Which is lrger A( ) or A()? 4 Grph of f 3 b. Which is lrger A() or A(4)? c. Where is A incresing? d. Eplin why A hs locl mimum t =. 3
25 Let G( ) f ( ) = 3 t dt, where the grph of the function f is given for 3 t 3.. Evlute G(3) nd G( 3). b. On wht intervl is G incresing? Grph of f c. On wht intervl is G concve down? d. Where does G hve mimum vlue?
26 Let f be continuous function over the intervl [,b ]. A function G is clled n ntiderivtive for f over the intervl [,b ] if G is continuous on [,b ] nd G' ( ) f ( ) ll in (,b ). Emple 4: Give few ntiderivtives for f ( ) =. = for Theorem: Fundmentl Theorem of Clculus Prt Let f be continuous function over the intervl [,b ]. If G is ny ntiderivvite for f over the b. intervl [,b ], then f ( ) d= G( b) G( )
27 Emple 5: Clculte the definite integrls using FTOC.. 5 d b. 3 d c. π/ cos d d. π/ 4 e d