# Math 1431 Section 6.1. f x dx, find f. Question 22: If. a. 5 b. π c. π-5 d. 0 e. -5. Question 33: Choose the correct statement given that

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1 Mth 43 Section 6 Question : If f d nd f d, find f 4 d π c π- d e - Question 33: Choose the correct sttement given tht 7 f d 8 nd 7 f d3 7 c d f d3 f d f d f d e None of these

2 Mth 43 Section 6 Are Under the Grph of Nonnegtive Function If y f is nonnegtive nd integrle over the intervl,, then the re under the curve y f over, is given y f d 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 elow: Theorem: If f is continuous on, nd if c For the function shown ove, Are of f ( d ) nd c Are of f ( d ) f( d ) c c c f d f d f d, then

3 Mth 43 Section 6 Emple : Given the grph of f, if the re of is nd re of is 8, find c f d Emple 6: Given f d, f d, the curve nd the -is from = - to = 4 4 f d 4 Find the re etween

4 Mth 43 Section 6 Other Properties of Definite Integrl Assume tht f nd g re continuous functions, where k is constnt numer kd k If f g over,, then f d g d 3 If m f M over,, then m f d M f d f d 4 If f is n odd function, then f d If f is n even function, then f d f d 3

5 Mth 43 Section 6 Try this one: Find the lower sum for f () =, [, ] if the prtition is 3 P,,,, 4 4 Try this one: Estimte 6 3 d y using left endpoint estimtes, where n = 6 Try this one: 3 6 Given f( ) d 3, f( ) d, f( ) d 9, find f( ) d 3 6 4

6 Mth 43 Section 6 Section 6: The Fundmentl Theorem of Clculus Recll from the previous section tht: For function f which is continuous on,, there is one nd only one numer tht stisfies the inequlity f, for ll prtitions P of L P I U P f, This unique numer I is clled the definite integrl (or just the integrl) of f from to nd is denoted y f ( d ) This numer cn e positive, negtive or zero Let f e continuous function over the intervl upper limit vries etween nd, Define new function y F f t dt Here, the F cn e seen s the re under the grph of f from F s the ccumulted re function If f hppens to e nonnegtive function, then to We cn think of So for emple, F f t dt if if then this is re under the function f from = to = F f t dt then this is re under the function f from = to =

7 Mth 43 Section 6 Emple : If f is the function whose grph is given elow, nd F f t dt, find F Fundmentl Theorem of Clculus Prt If f is continuous function over the intervl is continuous on, nd differentile on F f t dt, Note: If = then =,, then the function F f t dt d d, Moreover, F ' f t dt f, for ll in Emple : Let f e continuous function stisfying 3 f t dt Stte F() Find c Find F' f( ) F'' f '( ) Hence, where f is positive, F is incresing where f is negtive, F is decresing where f is zero, F hs possile m, min or inflection point where f is incresing, F is concve up where f is decresing, F is concve down

8 Mth 43 Section 6 Emple 3: Find '( ) F ( t t) dt F given Emple 4: Find d cos ( s) ds d Then find '(4 ) F d Emple : Find 3t dt d Recll the following definite integrl property from Section 6: f () d f() d

9 Mth 43 Section 6 Other times if the upper limit is not just, we ll need to use the chin rule Tht is, d v( ) f ( tdt ) f( v ( )) v'( ) d Emple 6: Find '( ) F given F dt t And yet, other times, if oth limits of integrtion re functions of we ll pply the following rule: d d v( ) u( ) f ( tdt ) f( v ( )) v'( ) f( u ( )) u'( ) Emple 7: Given 3 t dt, find F F' Let f e continuous function over the intervl, A function F is clled n ntiderivtive for f over the intervl, if F is continuous on, nd F' f for ll in, Emple 8: Give few ntiderivtives for f

10 Mth 43 Section 6 Theorem: Fundmentl Theorem of Clculus Prt Let f e continuous function over the intervl, If G is ny ntiderivvite for f over the intervl,, then f d GG Emple 9: Clculte the definite integrl d using FTOC

11 Mth 43 Section 6 F ( t 4 t ) dt Try this one: Let the function F e defined y 3 Find ny criticl numers for F d d Recll: F ' f t dt f Discuss the concvity of F

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# Math 1431 Section 6.1. f x dx, find f. Question 22: If. a. 5 b. π c. π-5 d. 0 e. -5. Question 33: Choose the correct statement given that

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1 Mth 43 Section 6 Question : If f d nd f d, find f 4 d π c π- d e - Question 33: Choose the correct sttement given tht 7 f d 8 nd 7 f d3 7 c d f d3 f d f d f d e None of these

2 Mth 43 Section 6 Are Under the Grph of Nonnegtive Function If y f is nonnegtive nd integrle over the intervl,, then the re under the curve y f over, is given y f d 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 elow: Theorem: If f is continuous on, nd if c For the function shown ove, Are of f ( d ) nd c Are of f ( d ) f( d ) c c c f d f d f d, then

3 Mth 43 Section 6 Emple : Given the grph of f, if the re of is nd re of is 8, find c f d Emple 6: Given f d, f d, the curve nd the -is from = - to = 4 4 f d 4 Find the re etween

4 Mth 43 Section 6 Other Properties of Definite Integrl Assume tht f nd g re continuous functions, where k is constnt numer kd k If f g over,, then f d g d 3 If m f M over,, then m f d M f d f d 4 If f is n odd function, then f d If f is n even function, then f d f d 3

5 Mth 43 Section 6 Try this one: Find the lower sum for f () =, [, ] if the prtition is 3 P,,,, 4 4 Try this one: Estimte 6 3 d y using left endpoint estimtes, where n = 6 Try this one: 3 6 Given f( ) d 3, f( ) d, f( ) d 9, find f( ) d 3 6 4

6 Mth 43 Section 6 Section 6: The Fundmentl Theorem of Clculus Recll from the previous section tht: For function f which is continuous on,, there is one nd only one numer tht stisfies the inequlity f, for ll prtitions P of L P I U P f, This unique numer I is clled the definite integrl (or just the integrl) of f from to nd is denoted y f ( d ) This numer cn e positive, negtive or zero Let f e continuous function over the intervl upper limit vries etween nd, Define new function y F f t dt Here, the F cn e seen s the re under the grph of f from F s the ccumulted re function If f hppens to e nonnegtive function, then to We cn think of So for emple, F f t dt if if then this is re under the function f from = to = F f t dt then this is re under the function f from = to =

7 Mth 43 Section 6 Emple : If f is the function whose grph is given elow, nd F f t dt, find F Fundmentl Theorem of Clculus Prt If f is continuous function over the intervl is continuous on, nd differentile on F f t dt, Note: If = then =,, then the function F f t dt d d, Moreover, F ' f t dt f, for ll in Emple : Let f e continuous function stisfying 3 f t dt Stte F() Find c Find F' f( ) F'' f '( ) Hence, where f is positive, F is incresing where f is negtive, F is decresing where f is zero, F hs possile m, min or inflection point where f is incresing, F is concve up where f is decresing, F is concve down

8 Mth 43 Section 6 Emple 3: Find '( ) F ( t t) dt F given Emple 4: Find d cos ( s) ds d Then find '(4 ) F d Emple : Find 3t dt d Recll the following definite integrl property from Section 6: f () d f() d

9 Mth 43 Section 6 Other times if the upper limit is not just, we ll need to use the chin rule Tht is, d v( ) f ( tdt ) f( v ( )) v'( ) d Emple 6: Find '( ) F given F dt t And yet, other times, if oth limits of integrtion re functions of we ll pply the following rule: d d v( ) u( ) f ( tdt ) f( v ( )) v'( ) f( u ( )) u'( ) Emple 7: Given 3 t dt, find F F' Let f e continuous function over the intervl, A function F is clled n ntiderivtive for f over the intervl, if F is continuous on, nd F' f for ll in, Emple 8: Give few ntiderivtives for f

10 Mth 43 Section 6 Theorem: Fundmentl Theorem of Clculus Prt Let f e continuous function over the intervl, If G is ny ntiderivvite for f over the intervl,, then f d GG Emple 9: Clculte the definite integrl d using FTOC

11 Mth 43 Section 6 F ( t 4 t ) dt Try this one: Let the function F e defined y 3 Find ny criticl numers for F d d Recll: F ' f t dt f Discuss the concvity of F

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Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type

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