Let's Integrate An Exponential Function

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[@ChefSalad]

You totally could have used the DI method with your first solution. The DI method doesn't require that the D column go to zero. Instead, you do all the diagonals and then the last row is an integral. Works like a charm.

[@SyberMath]

Thanks, Chef! 😍😉

[@Mehrdad_Basiry-fj4rl]

Fantastic... learn alot...❤❤❤.

[@SyberMath]

Glad to hear that ❤️

[@lesnyk255]

I never heard of the DI method before KZbin. I can see its advantages, but it strikes me as too mechanical - seems to require a lot of ritualistic motions. it gives you an answer, but loses track of why. "Here's the answer - now shutup." Longhand IBP might involve more steps, but hews more closely to 1st principles. Comes down to a matter of taste, I suppose. I prefer my old-timey u dv integration!

[@SyberMath]

Agree to a certain extent. If you know the mechanics, then using DI is a good shortcut. Imagine integrating x³eˣ with IBP. A lot of time could be saved. I think it's like long division vs. synthetic division (Horner's Method)

[@lesnyk255]

@@SyberMath Yeah, good point - kind of like memorizing the rules if differentiation vs. deriving them from scratch each time. Maybe it's just that I've been u-dv-ing all my life, and its become 2nd nature. Now that I'm retired and don't need to do math for anything but my own pleasure, DI just seems to take some of the fun out of it. Old dog, new trick!

[@calculus988]

You should do more integration by parts video

[@misterdubity3073]

I never heard of the DI method; looks like a shortcut for integration by parts. I like it; but it would be nice to see a verification that it works for a longer example, say integrating x^3 * e^x dx. Could be a segment of a future video.

[@SyberMath]

Great suggestion!

[@scottleung9587]

Very nice!

[@SyberMath]

Thank you! Cheers!

[@MASHabibi-d2d]

Thanks for an other video master...

[@SyberMath]

My pleasure!

[@kianmath71]

2sqrt(x)e^sqrt(x) - 2e^sqrt(x)

[@Alexey_Alex1]

ничего не понял, но очень интересно ))

[@SyberMath]

Спасибо! 😊

[@Alexey_Alex1]

@@SyberMath Eyvallah koç

[@RashmiRay-c1y]

Let t = sqrt(x) > x = t^2 and dx = 2tdt. So, I = 2\int dt t e^t = 2 \int dt t d/dt e^t = 2 t e^t - 2 \int dt dt/dt e^t = 2(t-1)e^t + C = 2(sqrt x -1) e^(sqrt x) + C.

[@lucafumagalli1829]

I solved it exactly the same way. Quite easy actually...

[@comdo777]

asnwer=1x isit

[@bjornfeuerbacher5514]

And yet again, you write a nonsensical answer. Do you like to make a public fool of yourself?