## anonymous 5 years ago represent ln5 as a definite intergral of the reciporcal function y=1/t

1. anonymous

$\int\limits_{1}^{5} 1/t$

2. anonymous

how would i grapth that to show the area it represents

3. anonymous

well you just need to graph 1/t and the area where x goes from 1 to 5

4. anonymous

how do i enter that in the calc

5. anonymous

with a graphic one you can do it. but I never used one

6. anonymous

do you know how 1/t looks like?

7. anonymous

no :/

8. anonymous
9. anonymous

the area you need is the one under the graph from x=1 to x=5

10. anonymous

so the first one?

11. anonymous

practice makes perfect, if you do calculus you need to be able to draw basic graphs of functions

12. anonymous
13. anonymous

so that is the the area? lol do i need to shade anything

14. anonymous

shade under the graph until the x axes

15. anonymous

perfect thanks

16. anonymous

what is the x-intercept of the tangent to y=e^x at x=45

17. anonymous

hmm, the derivative of e^x is e^x so the tangent at x=45 has slope of e^45 and the actual value is the same so then tangent line is y=e^45+e^45x the x intercept means that y=0 so 0=e^45+e^45x x=-1 that means that from x=45 you need to go back 1 to get the y intercept solution: x=44

18. anonymous

okay thats what i got. sweet

19. anonymous

well done!!