anonymous 5 years ago use the limit definition to evaluate the integral of (e^x-3)dx upper limit is 3, lower limit 1.

1. anonymous

let x-3 =t dx=dt hence our integral becomes $\int\limits_{1}^{3}e ^{t}dt$

2. anonymous

the question is Evaluate using the limit definition, $\int\limits_{1}^{3}(e^x-3)dx$

3. anonymous

maybe a mistake here. If you change from x - 3 to t you have to change the limits of integration. if x = 1 then t = 1-3 = -2 and if x = 3 then t = 3-3 = 0 so integral should be $\int^0_{-2} e^t,dt$

4. anonymous

oops Thought it was $e^{x-3}$ not $e^x-3$

5. anonymous

Are you really being asked to use a Reimann sum?

6. anonymous

yes

7. anonymous

is there a way to solve this without using reimann sum? do u have to take the natural log of e^x?

8. myininaya

int(e^x-3) =int(e^x)-int(3) =e^x-3x+C C is a constant

9. anonymous

hey could this be the final answer? u didnt use reimanns sum though

10. myininaya

well you also have the limits too this is alot easier than reimann sums it would be easy to do the reimann sums on int(3) but int(e^x) not sure never even tried to use riemann sum for that one

11. myininaya

int(e^x-3,x=3..1) =(e^x-3x,x=3..1) =(e^3-3(3))-(e^1-1(1)) thats the answer you can simplify of course

12. myininaya

oops mistake =(e^3-3(3))-(e^1-3(1))

13. myininaya

i have to go good luck

14. anonymous

thanks a lot!

15. anonymous

holy moly. i see no way to compute this reimann sum because we do not have a summation formula for this. the integral is easy enough: $\int^3_1e^x,dx=e^3-3$ $\int^3_1 3,dx=6$ since it is a constant and the length of the path is 2. so answer is $e^3-e-6$

16. anonymous

17. anonymous

sure. We want a definite integral. Easy way is to find the anti-derivative, plug in the upper limit, the lower limit, and subtract. It is easy to find the anti-derivative of $e^x$ since it is its own derivative. so $\int e^x,dx = e^x$ evaluate at 3 get $e^3$ evaluate at 1 get $e$ subtract and get $e^3-e$ now 3 is a constant. a horizontal line. So the integral is just base times hight. the hight is 3, the base is the length from 1 to 3 which is 2, and 3*2=6. I am ignoring the "-" because i am just going to subtract.

18. anonymous

i guess hight is spelled with an e as height. not that literate today. you could also find this integral by taking the anti-derivative: $\int 3 dx = 3x$ plug in 3 get 9. plug in 1 get 3. subtract get 6. but this is a waste of time, because the integral of a constant is always constant times length of path. $\int^b_a c dx=(b-a)c$

19. anonymous

thanks a lot

20. anonymous

no problem but i still have no idea how to compute this as a reimann sum

21. myininaya

see if this helps its a geometric series

22. myininaya

but maybe there is a mistake somewhere but i'm on the right track

23. myininaya

almost there

24. myininaya

use l'hospital's rule and you have $\lim_{u \rightarrow 0} \frac{ue^u}{1-e^{u}}=\lim_{u \rightarrow 0} \frac{e^u+ue^{u}}{-e^{u}}$

25. myininaya

$=\lim_{u \rightarrow 0} \frac{1+u}{-1}=\lim_{u \rightarrow 0} (-1-u)=-1-0=-1$

26. myininaya

and (-1)(e-e^3)=e^3-e :) and we win!