Improper Integrals Example 3 (KristaKingMath)

Improper Integral calculus example.

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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)

Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”

[kristakingmath]

@HazelnutFrappe I'm so glad they've helped you!! Thank you so much for all your support. :D

[kif]

you know what I like about you? even though it's a fairly simple problem, you go over every little detail so that we have a very solid foundation of the concept and details. if this were my current cal 2 professor, he would've rushed the problem without mentioning the little conceptual details that make the big difference.

[yens]

thanks for being everywhere in maths thanks so much

[kristakingmath]

Your welcome! :) Thank you so much for watching, and please let me know if there's anything else I can ever help with! :)

[kristakingmath]

that makes me so happy! i'm so glad it helped, good luck on your test!! :)

[kristakingmath]

You add C whenever you're dealing with an indefinite integral. If it's a definite integral, you don't need to add the C. Just to be clear, when you have the limits of integration, those little numbers at the top and bottom of the integral, you don't need to add C. If you don't have limits of integration, it's indefinite, and you need the C. Hope that helps! :)

[kristakingmath]

you can try it on your calculator to prove it to yourself! :) if you take e^(-10), you'll get about .00005. if you keep taking e to bigger and bigger negative numbers, that decimal will get smaller and smaller, that is, closer to 0. since we are taking the exponent to be an infinitely smaller and smaller negative number, we assume that we are going to get EXTREMELY close to 0, so we call the value 0. hope that helps! :D

[kristakingmath]

I'm so glad! Thanks for letting me know they're helping!! :D

[kristakingmath]

@subfourrun The exponent doesn't go to zero... the exponent goes to -infinity, which means that e raised to that exponent goes to 0. That leaves you with 10(0)=0, so 10-0=10. Plug e^-1000 into your calculator, and you'll see that it goes to 0. Hope that helps! :)

[kristakingmath]

@abhirup455 Yes. The only thing that makes it improper is the fact that the upper and/or lower limit of integration is infinite.

[skteboards]

much better than all the math teachers I've had for calc!

[kristakingmath]

i agree with you. i probably won't be making them anytime soon, because i'm not that good at it! but hopefully someone will make them soon!! :)

[zawar]

Thanks for the vid! Helps me understand my lecture notes. Keep 'em coming please!

[kristakingmath]

@shylildude this is one of my earlier videos...I've got the routine down now :)

[kristakingmath]

@kathryndelong wow, thanks. I am more than happy that I'm able to help. :)

[kristakingmath]

@kathryndelong Yes, that's exactly how to think about it! :)

[foofallsmilez]

calc 2 test this week yo have just saved my life :)

[Pearlie_e]

Thanks! Any chance you could record some examples of divergent improper integrals?

[mdfahd]

you are the best, i like the way u teach. and i like the way u look, coz i makes me understand faster ...

[mgouker]

Brittany, the first term is 10*(e^0) so 10. The second term is 10*(e^(-infinity)),
 which since the exponent is negative means that 10 is multiplied by something very near to zero (the limit *is* zero actually). Thus the second term evaluates to "almost zero" so you get 10-0=10. HTH :-)