Integration o Basic Fnctions Session 7 : 9/3
Antiderivation Integration Deinition: Taking the antiderivative, or integral, o some nction F(), reslts in the nction () i ()F() Pt simply: i yo take the integral o (), the reslting nction is (). Session 7 : 9/3
Relationship between dierention and integration d [ ] ( ) ( ) Dierentiation is the inverse o integration. '( ) ( ) C Integration is the inverse o dierentiation. **When the interval o integration is not speciied, called an indeinite integral. Session 7 : 9/3 3
Why Integrate? Common Uses:. To solve dierential eqations. To ind the area nder a crve or between crves Session 7 : 9/3 4
Notation: The integral o the nction ( ) 3 is written mathematically as ( ) ( ) 3 Integrand Integral symbol Dierential, i.e. With respect to Session 7 : 9/3 5
Eample: I ( ), then '( ) so, the integral o '( ) retrns the nction ( ) BUT Remember that the derivative o any constant is 0, so we have to accont or a possible constant when we integrate! Session 7 : 9/3 6
Remember Constants in Integration Integrate the ollowing nction: ( ) 4 ( ) ( 4 ) 4 C Where C is any constant Session 7 : 9/3 7
Basic Integration Rles. i k is a constant, k k C. i k is a constant, k ( ) k ( ) 3. 4. [ [ ( ) g ( )] ( ) g ( )] ( ) ( ) g ( ) g ( ) 5. n n n C, n Session 7 : 9/3 8
Eamples: Find the integral o: ( ) 4 g ( ) ( ) 3 4 ( ) Session 7 : 9/3 9
Finding Particlar Soltions Taking the integral, we are let with some nction () pls the constant C. Withot any rther inormation, we can t determine C, thereore ininite possibilities or the actal nction eist or any real vales o C. Session 7 : 9/3 0
Eample: '( ) F ( ) F ( ) C C C C0 so we have a general soltion, bt it isn t a particlar soltion becase we don t know what C is. C C Session 7 : 9/3
Particlar Soltions To ind a particlar soltion (i.e. the integral soltion at a certain vale o C), we need an initial condition An initial condition is a given vale o the integral soltion at any speciied vale o. Eample: '( ) 4 Find the general soltion (i.e. ind () withot determining C), then ind a particlar soltion (i.e. ind () and determine C) given the initial condition () 6. Session 7 : 9/3
Eample: Uses o Integration A model or the change in ozone concentration over time between 96 and 984 is given by: dc dt t 0 Where C is ozone concentration (ppm) and t is the elapsed time in years (96year 0). Given that in 964, the ozone concentration was 30ppm, determine the ozone concentration in 975. Session 7 : 9/3 3
Another Eample: The marginal cost o prodcing nits o a prodct is given by the eqation: dc 5 0. 0 Where C is the cost ($), and is the nmber o nits prodced. Given that prodcing nits o prodct costs $0, ind the cost o prodcing 300 nits. Session 7 : 9/3 4
The General Power Rle or Integration I is a dierentiable nction o (i.e. () is continos), then n d n d n C, n n Think reverse chain rle! Rewrite as ()d ( ( 4 ) 4 )( ) ( 4 ) d Session 7 : 9/3 5
Eamples: ( ) 3 ( 3 ) 3 Session 7 : 9/3 6
Maniplating Integrals ( 3 4 ) 4 ( ) 3 Session 7 : 9/3 7
Integration by Sbstittion Sometimes, terms to select or the integration process (i.e. () and d/) aren t obvios or aren t even there. In these cases, we can se integration by sbstittion to create a nction o within the eqation than can be integrated by the general power rle. Session 7 : 9/3 8
Eample Say we want to take the integral o the nction ( ) ( ) ( ) For this epression, there does not eist an obvios ()*d/ **Bt, we can transorm the integral to be in the orm ()*d/ Session 7 : 9/3 9
How? ( ) Step. Choose an obvios () (it s ine at this point that yo don t see a d/!). In this case, that is: d d Step. Take the derivative o with respect to. Write in the orm d/a. d Step 3. Solve or the epression as i the above eqation was an algebraic epression: Step 4. Sbstitte the epressions (in terms o and d) into the original integral and simpliy: ( ) d d Session 7 : 9/3 0
Session 7 : 9/3 Contining So now, we DO have an epression that has and d by sbstittion. This we know how to integrate sing the general power rle. d, n C n d n n Step 5. Use general power rle to integrate in terms o : D C C d 3 3 3 3 3 3 Why? Becase ½*C jst creates another constant, can keep calling it C or change to keep track Step 6. Resbstitte (remember we initially determined that ). So: ( ) 3 ) ( 3
Integration by Sbtittion: More Eamples 3 ( ) 5 ( ) Session 7 : 9/3
Eponential and Logarithmic Integrals Remember: Eponentials d ( e ) e Natral Logs d (ln( )) d ( e C ) e d (ln( ) C ) d d ( e ) e ( d ) d Where is a nction o (ln( )) d Session 7 : 9/3 3
Integrating Eponentials e e C e d e d e C Session 7 : 9/3 4
Eamples: e e ( ) Original nction to integrate with respect to e C e d Rewrite, i possible, in orm e d Soltion 4e Session 7 : 9/3 5
Integrating Logarithmic Fnctions ln C d d ln C Where is a nction o Session 7 : 9/3 6
Eamples Step. Recognize i and d/ eist in epression Step. Rewrite to inclde only and d Step 3. Use rles rom previos slide to ind integral soltion 4 3 3 5 Session 7 : 9/3 7
Rewriting Integral Eamples: e Session 7 : 9/3 8
Tomorrow: Integration by Parts Deinite Integrals Solving Dierential Eqations by Integrating Improper Integrals Session 7 : 9/3 9