Calculus 1 Lecture 4.4: The Evaluation of Definite Integrals

Calculus 1 Lecture 4.4: The Evaluation of Definite Integrals

[andreas_iced]

internet audience: "thank you so much, professor leonard!! you literally saved my grade in engineering math 😭"
his students, specifically that _one dude_ : "yeah, i did your way, but i got no results"

[darthboganis]

The Integral 0:30
> The Area Formula restated 1:10
Definite Integrals 4:50
> geometrical interpretation 7:00
> Properties 16:00
>w/ Example 27:00

Note: Riemann Sums are the addition of n number of rectangles to approximate the area. Taking the limit of the sum adds an infinite number of rectangles, which gives the area. This is equivalent to taking the integral of the function. A greater n gives a closer approximation of the true area.

[Mcinema]

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[liliantu]

you have been my calc 1 teacher so far

[cus]

Hey Prof Leonard, I just wanted to that your calc 2 video series will be the only reason I will pass this course. Your videos are still helping students succeed even 9 years after the videos were posted!

[techconstellation]

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youtube. Everything was perfectly clear and easy to follow along!

[jtwoms]

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[ayushsingh]

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[alie]

Truly making university calculus simple. Ciao

[mgrace]

OSU need professors like you! thanks for teaching and sharing your lectures.

[gabrielrequena]

Jesus christ, you are jacked as hell. Besides Calc man... How did you get so big?

[djtfan]

Area of triangle @ 10:50 is wrong, it should be (3*3)/2 and not (3*2)/2. Class is really not paying attention :P


EDIT: ofc he fixes it seconds later...

[kmfdmboots]

Thanks, for this video. Made my homework 40 times easier

[omarf]

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[Alya]

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[dhuratabali]

You explain very well but in properties 4 you can use bracket it is very important.

[OwaisJaffary]

Thank you sir for your incredible effort... truly a mesmerizing teacher

[giorgipachuashvili]

you are, ,The Teacher ! '' professor.

[adrianloh]

Finally, i heard he said "Riemann sum"...in this video....

[TehBluPanda]

So..., since the negative of an area implies the area under it (or a reflexion across the x axis) does that mean that the area implies a position on the graph, rather than being indicative of the total space inside an interval? (I'm not too knowledgable on what areas are either, so bear with me)