Understand Application of Integrals with the help of numerical. click on the link to Watch the VIDEO explanation: Watch Video
Application of Integrals - Problem 1
The Problem
Find the area enclosed between the parabolas y square is equal to 4ax and x square is equal to 4ay.
The Solution:
Points of intersection of the curves can be obtained by solving the equations.
y square is equal to 4ax and x square is equal to 4ay
Therefore, x is equal to 0 and x is equal to 4a.
Area enclosed between the given curves
A is equal to integral zero to 4a F1 of x minus F2 of x dx.
Therefore, the area is equal to 16a square by 3 square units.
Application of Integrals - Problem 1
The Problem
Find the area enclosed between the parabolas y square is equal to 4ax and x square is equal to 4ay.
The Solution:
Points of intersection of the curves can be obtained by solving the equations.
y square is equal to 4ax and x square is equal to 4ay
Therefore, x is equal to 0 and x is equal to 4a.
Area enclosed between the given curves
A is equal to integral zero to 4a F1 of x minus F2 of x dx.
Therefore, the area is equal to 16a square by 3 square units.
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