calculus help: find the arc length: application of integration: 6xy=x^4 3 from x=1 to x=2
Published 3 years ago • 899 plays • Length 3:26Download video MP4
Download video MP3
Similar videos
-
5:39
calculus help: find the arc length by integration - technique how to solve
-
3:18
calculus help: find the arc length f(x)=2 x^(3/2) , 0≤x≤1 - techniques - solutions
-
9:36
calculus help: find the arc length of the circle by integral (x-1)^2 (y-3)^2=16 - integration
-
4:30
calculus help: arc length - x=y^4/4 1/(8y^2 ) , 1≤y≤2 - application of integrations - solved!!!
-
3:49
calculus help: find the arc length of the curve y=1/8 (-x^2 8lnx) from x=4 to x=6 - integration
-
3:57
calculus help: the arc length and how to answer - techniques
-
2:43
florel trick by priya ma'am ❤️
-
13:59
4 curves to save your math grades
-
16:21
find the arc length of a function with calculus
-
![calculus help: find the arc length y=4/3 x^(3/2) on the interval [0,1] - integration](https://i.ytimg.com/vi/P-oL9cxDCF8/mqdefault.jpg)
2:28
calculus help: find the arc length y=4/3 x^(3/2) on the interval [0,1] - integration
-
22:37
calculus help: application of parabola and integration (arc length) - ballistic missile - red alert
-
12:31
calculus: application of integration: find velocity and position, volume rotate, area, arc length
-
3:36
determine the area of the region bounded by the curve y=-x^3-x, the x-axis and the line x=-1 and x=2
-
![calculus help: determine the arc length of f(x)=(x/2)^(3/2) of [0;2] - techniques - integrations](https://i.ytimg.com/vi/t-f8D9TKxV4/mqdefault.jpg)
2:58
calculus help: determine the arc length of f(x)=(x/2)^(3/2) of [0;2] - techniques - integrations
-
2:31
application of integration: find the area between the curve g(x)=x^3- 0.25x^4 and the line f(x)=2x-8
-
5:48
calculus help: find the length of the curve given by y=lnx from x=1 to x=e
-
4:41
calculus help: arc length - the lengths of the following curves - x=(y/7)^2-49/8 ln(y/7) ,6≤y≤30
-
30:47
arc length calculus problems,
-
3:03
calculus help: find the arc length of the function f(x)=2/3 (x^2 1)^(3/2) on the interval 0≤x≤1
-
5:00
arc length - integration: find the length of the curve y=(4-x^(2/3) )^(3/2) from x=1 to x=8
-
4:31
calculus help: find the arc length - y=1/8 (-x^2 8 ln(x) ) from x=2 to x=6 - integration
-
5:35
calculus help: find the length of the curve y=arcsin(e^(-x) )from x=0 to x=1 - techniques