Definite Integrals

[Ineptskills]

Hi,

I am working on some definite integrals, however, my mathematics ability is not top notch.

Could someone please explain how to evaluate a definite integral in the simplest way possible?

I have attached and example to this thread

Thanks everyone

** I can't seem to type the example. I can describe it though:

It is an 'S' with a '2' and a '0' below next character is 'e' to the power of '2x' and it finishes with 'dx'

i believe the task is to evaluate between 0 and 2 on a graph with a function of e to the power of 2x

I look at other examples and I am just lost. any help? please

thanks

 

[stapel]

All I'm seeing is a gray-splotched square. Please reply with the typed-out text of the exercise.

If you're not sure how to format math as typed text, please review this article for information. Thank you!

 

[Ineptskills]

All I'm seeing is a gray-splotched square. Please reply with the typed-out text of the exercise.

If you're not sure how to format math as typed text, please review this article for information. Thank you!

Hi,

Thanks, I'll check it out and repost

Thanks again,

 

[Ishuda]

...i believe the task is to evaluate between 0 and 2 on a graph with a function of e to the power of 2x ...

Going by what you say, I believe it would be
$\displaystyle \int_0^2\, e^{2x}\, dx$

In this case you would need the function whose derivative is e^2x. Call that function F(x) and the answer would then be
$\displaystyle \int_0^2\, e^{2x}\, dx\, =\, F(2)\, -\, F(0)$

BTW:Just in case you would care to know, that last equation was created using LaTex and looks like

[tex]\int_0^2\, e^{2x}\, dx\, =\, F(2)\, -\, F(0)[/tex]

The \, are just spaces to make it look better (IMO). You can find out more about Latex at
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
which has some links to even more details.

 

[Ineptskills]

Going by what you say, I believe it would be
$\displaystyle \int_0^2\, e^{2x}\, dx$

In this case you would need the function whose derivative is e^2x. Call that function F(x) and the answer would then be
$\displaystyle \int_0^2\, e^{2x}\, dx\, =\, F(2)\, -\, F(0)$

BTW:Just in case you would care to know, that last equation was created using LaTex and looks like

[tex]\int_0^2\, e^{2x}\, dx\, =\, F(2)\, -\, F(0)[/tex]

The \, are just spaces to make it look better (IMO). You can find out more about Latex at
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
which has some links to even more details.

Hi Ishuda,

Thanks very much for your help. I have been watching videos on youtube about integrals and definite integrals. I kept thinking I had to somehow graph this in order to evaluate it.
Thanks again :cool:

and thanks for the advice about LaTex too. I'll check it out

 

[Steven G]

Going by what you say, I believe it would be
$\displaystyle \int_0^2\, e^{2x}\, dx$

In this case you would need the function whose derivative is e^2x. Call that function F(x) and the answer would then be
$\displaystyle \int_0^2\, e^{2x}\, dx\, =\, F(2)\, -\, F(0)$

BTW:Just in case you would care to know, that last equation was created using LaTex and looks like

[tex]\int_0^2\, e^{2x}\, dx\, =\, F(2)\, -\, F(0)[/tex]

The \, are just spaces to make it look better (IMO). You can find out more about Latex at
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
which has some links to even more details.

I especially like the spaces.

 

[Steven G]

Hi,

I am working on some definite integrals, however, my mathematics ability is not top notch.

Could someone please explain how to evaluate a definite integral in the simplest way possible?

I have attached and example to this thread

Thanks everyone

** I can't seem to type the example. I can describe it though:

It is an 'S' with a '2' and a '0' below next character is 'e' to the power of '2x' and it finishes with 'dx'

i believe the task is to evaluate between 0 and 2 on a graph with a function of e to the power of 2x

I look at other examples and I am just lost. any help? please

thanks

You should know that the integral of (e^u)du = e^u + c.
So do a u substitution with u=2x and all will work out nicely.

 

[Ineptskills]

You should know that the integral of (e^u)du = e^u + c.
So do a u substitution with u=2x and all will work out nicely.

Hi Jomo,

I think I may have done it. I attached a file with what I hope is the correct answer. Do you think you could check it?

Thanks

Ineptskills

 

[Steven G]

Hi Jomo,

I think I may have done it. I attached a file with what I hope is the correct answer. Do you think you could check it?

Thanks

Ineptskills

Nicely done. You can always check your work by taking the derivative of your answer and see if you get back the integrand. (d/dx)(.5e^(2x) + c) does equal e^(2x)

 

[Ineptskills]

Nicely done. You can always check your work by taking the derivative of your answer and see if you get back the integrand. (d/dx)(.5e^(2x) + c) does equal e^(2x)

Thanks very much! I have never done any calculus before now and it's not coming to me very easily!

The Darwin quote you sign your messages with sums me up nicely apart from the mathematician part!

 

[juragan]

This seems like a very interesting thread. new Member permission join this discussion

 

[Ishuda]

This seems like a very interesting thread. new Member permission join this discussion

Hi juragam,
Welcome. Sure, jump right in. There are some general 'rules' about posting but basically they are 'play nice with the other kids'. That doesn't mean though that Jomo, Denis, Subhotosh Khan, or actually anyone can't tell me to go stand in the corner if I make a dumb mistake.

If you have a question about a problem the rule is basically
What have you done so far? Please show us your work so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

So, once again, welcome and jump right in.

 

[Deleted member 4993]

Hi Jomo,

I think I may have done it. I attached a file with what I hope is the correct answer. Do you think you could check it?

Thanks

Ineptskills

There are some problems with the solution you have shown.

$\displaystyle \displaystyle{\int_0^2 e^{2x}dx}$

Substitute

u = 2x ..... Now you need to change the limits also

du = 2 dx

$\displaystyle = \ \displaystyle{\int_0^4\frac{1}{2} e^{u}du}$

$\displaystyle = \ \displaystyle{\frac{1}{2} \left [e^{4} \ - \ 1 \right ]}$

This is a definite integral - so no constant of integration!