Find the area enclosed by the line y=4x and the curve y=x^2?

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Answer 1 · 2017-07-17T15:23:38

#A = 32/3# square units

Start by finding the intersection points. These will determine our bounds of integration.

#x^2 = 4x#

#x^2 - 4x = 0#

#x(x -4) = 0#

#x= 0 or 4#

We now determine which curve lies above which, because we want to find #A_("above curve") - A_("below curve")#.

If we test a point in the interval #(0, 4)#, say #3#, we get

#y_(x = 3) = 4(3) = 12#

#y_(x = 3) = 3^2 = 9#

So the linear graph will be above the quadratic graph in the interval of area we are trying to find.

We now set up our integral.

#A = int_0^4 4x - x^2dx#

We separate the integrals.

#A = int_0^4 4xdx - int_0^4 x^2 dx#

#A = [2x^2 - 1/3x^3]_0^4#

We evaluate using the 2nd fundamental theorem of calculus.

#A = 2(4)^2 - 1/3(4^3) - (2(0)^2 - 1/3(0)^3)#

#A = 32 - 64/3#

#A = 32/3#

And don't forget that area is in square units.

Hopefully this helps!