Calculus 2 - Integration: Finding the Area Between Curves (5 of 22) Ex. 5: y=(1/4)x^2, y=x

[@tiberiusG]

i can't adequately express in words my gratitude for your videos, or praise for their quality.

[@harrasmansoor4098]

Thankyou so much for clearing my concepts ❤

[@MichelvanBiezen]

Glad you find our videos helpful.

[@edzifer]

Thanks, I am hoping an exercise like this comes to my exam

[@MasayoMusic]

Thank you.

[@sk_ismail_hsn]

I'm from Bangladesh.Thank you sir

[@MichelvanBiezen]

Welcome to the channel!

[@sydniadams7318]

I took my dA horizontally and got 40/3. Please tell me what I did wrong. After integrating, I was left with [8y^(3/2)/3] - [y^2/2] from 0 to 4. I know this way seems much harder but in the end they should compute the same answer, correct?

[@MichelvanBiezen]

Why not use the technique shown in the video?

[@sydniadams7318]

Once I saw the equations, I hit pause to attempt the problem on my own. I ended up doing it horizontally but when I hit play I saw that you did it vertically. Either way I should have gotten the same answer is my point.

[@billwindsor4224]

@Syndi Adams - your overall statement is correct that you should get the same answer by calculating in either the dx (vertical) direction, or the dy (horizontal) direction. But when you calculate your integral, you are integrating the term: [2 * y^(1/2) - y ] dy , which integrates to: [ (4/3) * y^(3/2)] - (1/2) * y^2 , evaluated at the limits y=4 and y=0, which you will find also equals 8/3 as Michel shows.

[@wanderleyapparecidovieira2282]

Very well explained,but a little messy...

[@anthonysenju4044]

Thanks