These videos are very helpful. Thank you so much for taking the time to make them - I'm studying calculus after 6 years in between cal 2 and cal 3 and this is very helpful. Thanks again, Jennifer
Great Video... You helped me pass my test. Especially when talking about the boundaries and when to integrate y or x first. Other videos don't discuss this. Thanks sooo much! :-)
Great video, as usual. Amazing explanation. Just a very small trivial note at time 6:14. The y integral limits are from x^3 to sqrt(x), that's 100% correct, no doubt about it. But then, the x integral limits, you gave those for y to be from 0 to 1, and this should be instead the integral limits of x from 0 to 1, you've said, the y limits from 0 to 1. I know this sometimes could be confusing since in this example, both integral limits for y looks like the same integral limit for x which goes from 0 to 1.
Yea, this got me at first of solving these type of problems. At time 2:20, this point is called intersection points. To find them, set both boundary functions to equal each other. Since, y=x-1; y=(1/2)x so set them equal to each other, x-1=(1/2)x => x-(1/2)x=1 => (1/2)x=1 => x=2, plug 2 in y=x-1 or y=(1/2)x you get y=1, so there is only one intersection point, i.e. (2,1). You can apply the same logic to the second example.
These videos are very helpful. Thank you so much for taking the time to make them - I'm studying calculus after 6 years in between cal 2 and cal 3 and this is very helpful.
Thanks again,
Jennifer
Your videos are so awesome. I rather spend 3hrs at home study from your video than wasting time going to classes which all I do is fall asleep.
first time to watch this but everything is already clear...
One of the best videos !!
Thank you for the comment.
Great Video... You helped me pass my test. Especially when talking about the boundaries and when to integrate y or x first. Other videos don't discuss this. Thanks sooo much! :-)
Great video, as usual. Amazing explanation. Just a very small trivial note at time 6:14. The y integral limits are from x^3 to sqrt(x), that's 100% correct, no doubt about it. But then, the x integral limits, you gave those for y to be from 0 to 1, and this should be instead the integral limits of x from 0 to 1, you've said, the y limits from 0 to 1. I know this sometimes could be confusing since in this example, both integral limits for y looks like the same integral limit for x which goes from 0 to 1.
Very helpful and save quite alot of time. Thank you so much.
Thanks for your videos; they're lifesavers!
how do you get the point (2,1)? I have problem knowing at which points they cross each other
Yea, this got me at first of solving these type of problems. At time 2:20, this point is called intersection points. To find them, set both boundary functions to equal each other. Since, y=x-1; y=(1/2)x so set them equal to each other, x-1=(1/2)x => x-(1/2)x=1 => (1/2)x=1 => x=2, plug 2 in y=x-1 or y=(1/2)x you get y=1, so there is only one intersection point, i.e. (2,1). You can apply the same logic to the second example.
great video!!! keep the good job!
With respect to, not with respects to.
did u make a mistake? -(0-1/12(-1)^4= -(1/12) there it's 1/6-1/12-1/12= 1/6-2/12= 1/12
oh never mind i see i made the mistake