Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is sometimes clled the nti-derivtive, but is more commonl known s the integrl, nd is given the sign. If f() = n, then n d is the integrl of n with respect to Indefinite Integrls nd the Constnt of Integrtion Consider the three functions () = 2 + 2 + 5, b() = 2 + 2 8 nd c() = 2 + 2 -. 4 In ech cse, the derivtive of the function is the sme, i.e. 6 + 2. This mens tht (6 2)d hs more thn one nswer. Becuse there is more thn one nswer, we s tht this is n indefinite integrl, nd we must include in the nswer constnt vlue C, to represent the 5, -8, etc which we would 4 need to distinguish () from b() from c ( ) etc. To find the integrl of function, we do the opposite of wht we would do to find the derivtive: f() Multipl b the old power Decrese the power b f () f() f () n n In generl: d C (n ) n IN tegrtion IN creses the power!. Write s n 2. Increse the power b. Divide b the new power I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 4 of 9
Emple : Find (remember +C ): ) 2 d b) 4 t 2 dt c) ( 5 4) d d) dg g 4 4 (g 0) 5 e) 6 p dp f) 2 d ( 0) The Definite Integrl A definite integrl of function is the difference between the integrls of f( ) t two vlues of. Suppose we integrte f( ) nd get F( ). Then the integrl of f( ) when = would be F(), nd the integrl when = b would be F(b). The definite integrl of f( ), with respect to, between nd b, is written s: b f( ) d F(b) F() (where b > ) For emple, the integrl of f( ) = 2 2 4 between the vlues = - nd = 5 is written s 5 2 (2 4)d nd reds the integrl from - to 5 of 2 2 4 with respect to. Note: definite integrls do NT include the constnt of integrtion! Emple 2: Evlute (2 ) d b f() [F(b) C] [F() C] F(b) F() To find definite integrl: prepre the function for integrtion integrte s norml, but write inside squre brckets with the limits to the right sub ech limit into the integrl, nd subtrct the integrl with the lower limit w from the one with the higher limit I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 5 of 9
Emple : Evlute p p dp 2 0 Are Between Grph nd the is. In the digrm opposite, the re of the shded section cn be obtined b finding the re under the grph from to b, nd subtrcting the re from to. =f( ) The vlue of ech of these res cn be determined b integrting the function nd substituting b or respectivel. b The re enclosed b the curve = f(), the lines =, = b nd the is is equl to the definite integrl of f() between nd b i.e. Are = b f( ) d Emple 4: For ech grph below, (i) write down the integrls which describe the shded regions (ii) clculte the re of the shded region ) b) = 6 = 2 9 I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 6 of 9
NTE: Emple 4b shows tht res UNDER the is give NEGATIVE vlues! Emple 5: 7 ) Evlute (2 6) d b) (i) Sketch below the re described b the 7 integrl (2 6 ) d. The nswers for 5 nd 5b do not mtch! This is becuse the re below the is nd the re bove cncel ech other out (s in 4b, res below the is give negtive vlues). When clculting the re between curve nd the is:. Determine the limits which describe the sections bove nd below the is 2. Clculte res seprtel. Add together, IGNRING THE NEGATIVE VALUE F THE SECTIN BELW THE AXIS. Emple 6: Determine the re of the regions bounded b the curve = 2 4 + nd the nd es. = 2 4 + I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 7 of 9
Are Between Two Curves Consider the re bounded b the curves = ( 2) 2 nd =. =( 2) 2 = =( 2) 2 = =( 2) 2 = Are = 2 d - ( 2) d The digrms bove show tht the re between the curves is equl to the re between the top function ( ) nd the is MINUS the re between the bottom curve (( 2) 2 ) nd the is. =g( ) b =f( ) The re between the curves = f( ) nd = g( ) (which meet t the points where = nd = b ) is given b: where: A b (f( ) g( )) d f( ) is the TP function nd g( ) is the BTTM b > Emple 7: Write down the integrls used to determine the res shown below: ) b) c) = 6 2 = 5 = 2 2 = = 2 = I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 8 of 9
To find the re between two curves:. Mke sketch (if one hs not been given) 2. Find points of intersection (mke = nd solve). Subtrct the bottom function from the top function, PUTTING THE BTTM FUNCTIN IN BRACKETS! 4. Integrte Emple 8: Find the re enclosed between the curve = 2 5 nd the line = = = 2 5 I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 9 of 9
Differentil Equtions If we know the derivtive of function (e.g. f ( ) = 6 2 ) insted of the function itself, we cn obtin the function b integrtion. This is clled differentil eqution, nd gives us the function in terms of nd C (which we cn then evlute if we hve point on the grph of the function). Emple 9: The grdient of tngent to the curve of = f( ) is 24 2 + 0. Epress in terms of, given tht the grph of = f( ) psses through the point (-, -0). Pst Pper Emple: Find re enclosed between the curves = + 0 2 2 nd = + 5 2. = + 0 2 2 = + 5 2 Integrtion: Topic Checklist Topic Questions Done Help? Finding indefinite integrls C Eercise 9H, p 64 (ll) Y/N Y/N Eercise 9I, p 65, Q ( n) Y/N Y/N Definite Integrls C Eercise 9L, p 69, Q Y/N Y/N Eercise 9K, p 67, Q Y/N Y/N C Are under curve Eercise 9N, p 72, Q, (bove nd below ) Y/N Y/N A/B Eercise 9N, p 72, Q 4; Eercise 9P, p 75, Q 5 Y/N Y/N Are between two grphs C Eercise 9P, p 74, Q Y/N Y/N A/B Eercise 9P, p 74, Q 2, 4; Eercise 9R, p 78, Q 7, Y/N Y/N Differentil Equtions C Eercise 9Q, p 76, Q 2, Y/N Y/N A/B Eercise 9R, p 79, Q 4, 5 Y/N Y/N I Brson, mended b M Dorn & E Mwell: Airdrie Acdem Mthemtics Deprtment, 20/4 Pge 40 of 9