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Integration with infinity limit

Find definite integral _-∞ʃ ⁰ e^x dx

I've tried looking up how to do this but can't find anything that makes sense to me... can't seem to find anything about it being - ∞ either

Reply 1

TeeEm

Original post by jacksonmeg

Find definite integral _-∞ʃ ⁰ e^x dx

I've tried looking up how to do this but can't find anything that makes sense to me... can't seem to find anything about it being - ∞ either

have you seen improper integrals in your lessons/lectures?

Reply 2

davros

Original post by jacksonmeg

Find definite integral _-∞ʃ ⁰ e^x dx

I've tried looking up how to do this but can't find anything that makes sense to me... can't seem to find anything about it being - ∞ either

Well, what do you get when you work out the indefinite integral? What's the value of that function at x = 0? What's value does the function approach as

x \rightarrow -\infty

? So what's the overall answer?

Reply 3

jacksonmeg

Original post by TeeEm

have you seen improper integrals in your lessons/lectures?

No... we had 1 lecture covering chain rule, product rule, quotient rule, partial differentiation, integration and taylor series in 50 mins :/

Reply 4

jacksonmeg

Original post by davros

Well, what do you get when you work out the indefinite integral? What's the value of that function at x = 0? What's value does the function approach as

x \rightarrow -\infty

? So what's the overall answer?

x=0 then the integral is equal to 1

When it approaches infinity does it become 0?

Reply 5

Smaug123

Original post by jacksonmeg

Find definite integral _-∞ʃ ⁰ e^x dx

I've tried looking up how to do this but can't find anything that makes sense to me... can't seem to find anything about it being - ∞ either

Completely as an aside (ignore if you've not done probability before, and ignore if you don't immediately understand it), you might be able to recognise this one with no thought at all because the exponential distribution with mean

\frac{1}{\lambda}

has PDF

\lambda e^{- \lambda x}

Reply 6

TeeEm

Original post by jacksonmeg

No... we had 1 lecture covering chain rule, product rule, quotient rule, partial differentiation, integration and taylor series in 50 mins :/

wow!

ok.

replace -∞ with a letter, say h.
carry out the integration, getting an answer in terms of h

Reply 7

jacksonmeg

Original post by TeeEm

wow!

ok.

replace -∞ with a letter, say h.
carry out the integration, getting an answer in terms of h

I know q-q nobody told me a chemistry degree would have some much maths ...

I ended up with 1 - e^h

(edited 8 years ago)

Reply 8

davros

Original post by jacksonmeg

x=0 then the integral is equal to 1

When it approaches infinity does it become 0?

Correct (well you meant minus infinity, not infinity)!

So putting the two bits together gives you....?

Reply 9

Smaug123

Original post by jacksonmeg

No... we had 1 lecture covering chain rule, product rule, quotient rule, partial differentiation, integration and taylor series in 50 mins :/

To write down what TeeEm said formally (in a way which might match what you've been lectured):

We define

\int_{-\infty}^a f(x) dx

to be

\displaystyle \lim_{h \to -\infty} \int_{h}^a f(x) dx

, if the limit exists.

Reply 10

TeeEm

Original post by jacksonmeg

I know q-q nobody told me a chemistry degree would have some much maths ...

I ended up with 1 - e^h

now let h tend to -∞

do you know what this gives?

EDIT: I am out... Too many helpers here.

Reply 11

jacksonmeg

Original post by davros

Correct (well you meant minus infinity, not infinity)!

So putting the two bits together gives you....?

Reply 12

davros

Original post by jacksonmeg

Correct

(and I see someone's given you a formal definition of how to evaluate these integrals above)

Reply 13

Smaug123

Original post by TeeEm

EDIT: I am out... Too many helpers here.

I'M HELPING
I PROMISE I'M USEFUL
PLEASE TELL ME I HELPED

Sorry for driving you away :colondollar:

my fault!

Reply 14

TeeEm

Original post by Smaug123

I'M HELPING
I PROMISE I'M USEFUL

Sorry for driving you away :colondollar:

no worries
I am very busy I promise, so you are doing me a big favour (Davros too)

Reply 15

jacksonmeg

Original post by davros

Correct

(and I see someone's given you a formal definition of how to evaluate these integrals above)

yay

thanks for all your help!

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