how to find the area? (1 Viewer)

[bubb]

For any point p on the curve y= x^3 . Prove that the area under the curve is one quarter of the area of the rectangle ?

I went this far into the question

integrating between p and 0 y^1/3 dy = (3(p^4/3))/4

and is the area of the rectangle p^2?

 

[zhertec]

For any point p on the curve y= x^3 . Prove that the area under the curve is one quarter of the area of the rectangle ?

I went this far into the question

integrating between p and 0 y^1/3 dy = (3(p^4/3))/4

and is the area of the rectangle p^2?

Integrate y=x^3 dx where the boundaries are x=p and x=0
then you get x^4/4 (no plus c as it is definite integral )
sub p in to get p^4/4 which is area under curve.
area = l x b
l = x
B = y

hence to get y sub p into x^3
Therfore your final asnwer for rectangle area will be p^4

Finally area under curve/area rectangle = (p^4/4)/p^4 which equals 1/4

 

[Flop21]

What is "the rectangle"?