Integration of Exponential Functions (1 Viewer)

[Dragie]

Could someone simplistically explain the method of integrating an exponential? I've tried to understand and I've read a lot of sources but it wont sink in. Would anyone care to explain? Thank you so much!

 

[insert-username]

If the derivative of e^x is e^x, then it follows that its integral is also e^x. This, along with the chain rule is all you probably need to know to integrate exponential functions:

∫e^2x+1dx

If we finish with e^2x+x, we must have started with e^2x+1, since the derivative of e^x is e^x. However:

d/dx(e^2x+1) = 2e^2x+1 (chain rule)

This is double what we want to end up with, so we add a half to balance the coefficients:

∫e^2x+1dx

= 1/2e^2x+1

Just remember that when you integrate you need to end up with something that differentiates to get the original function.


I_F

 

[Dragie]

Hey thank you so much to both of you but I just need help with one more question:


1. Evaluate this definite integral:
1
∫ e^3x dx
0

2. Evaluate this definite integral:
2
∫ 1/2(e^x + e^-x)dx
0

I'm slowly getting it but I think I need a bit more...

 

[pLuvia]

1
∫ e^3x dx
0

= [1/3*e^3x ] Stick in the limits and you should get answer

2
∫ 1/2(e^x + e^-x)dx
0

= 1/2 [ e^x - e^-x ] Stick in the limits and you should get the answer

 

[Dragie]

1
∫ e^3x dx
0

= [1/3*e^3x ] Stick in the limits and you should get answer

2
∫ 1/2(e^x + e^-x)dx
0

= 1/2 [ e^x - e^-x ] Stick in the limits and you should get the answer

Oh is that it? Haha I've been killing myself over integration of these functions for a few days and I just got it! Lol damn I'm slow sometimes.

 

[Dragie]

2 more questions I promise!! How would these be done:

1.
x^2.e^({2x^3}+1)

2.
2x/(e^X^2)


It's really hard to type these out - does anyone know how to superscript on here

 

[pLuvia]

I assume integrate them?

∫ x²e^2x3+1 dx
Let u=2x³+1
du = 6x² dx

=1/6 ∫ 6x²e^u dx
=1/6 ∫ e^udu
=1/6[e^u]
=1/6*e^2x3+1+C

∫ 2x/(e^x2)
Let u=x²
du=2xdx
= ∫ 2x/e^udx
= ∫ du/e^u
= [-e^-u]
= -1/e^x2+C

To use superscript use this code

[sup ]Insert Text[/sup ]

With subscript

[sub ]Insert Text[/sub]

 

[Riviet]

General rule:

∫f'(x)e^f(x) dx
let u=f(x)
du/dx = f'(x)
du = f'(x) dx
substitute u and du into your original integral:
∫f'(x)e^f(x) dx = ∫ e^u du
= e^u + C

Remember to manipulate constants to get your integral into the form f'(x)e^f(x)

 

[Dragie]

Thank you to all of you - I'm actually not lost anymore

 

[Rax]

Differentiating exponentials, You differentiate the power and multiply it by the original exponential

Integrating exponentials, Differentiate the power and DIVIDE the original exponential by the derivative

Done and Done

 

[imranium]

I assume integrate them?

∫ x²e^2x3+1 dx
Let u=2x³+1
du = 6x² dx

=1/6 ∫ 6x²e^u dx
=1/6 ∫ e^udu
=1/6[e^u]
=1/6*e^2x3+1+C

∫ 2x/(e^x2)
Let u=x²
du=2xdx
= ∫ 2x/e^udx
= ∫ du/e^u
= [-e^-u]
= -1/e^x2+C

To use superscript use this code

[sup ]Insert Text[/sup ]

With subscript

[sub ]Insert Text[/sub]

----------------------------------


wooww.. hold on a sec. so where does the x^2 from (∫ x²e^2x³+1 dx) go?????????????????????????

 

[lolokay]

they substituted in 6x²dx = du

 

[dakiu]

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