The area of the region bounded by the curve y=x^2 and y=4x-x^2 is ?

Answer 1

#color(blue)[A=int_0^2(4x-x^2)-(x^2)*dx=8/3=2.667 (unite)^2]#

Firstly lets find the limits of the Area

since #y=x^2# and #y=4x-x^2#

#x^2=4x-x^2#

#2x^2-4x=0 rArr 2x*(x-2)=0#

#2x=0 rArr x=0#

since #x=0 rArr y=x^2 rArr y=0#

#A=(0,0)#

#x-2=0 rArr x=2#

since #x=2 rArr y=x^2 rArr y=4#

#B=(2,4)#

if we take dx slice (that mean the Area revolving about x-axis)

The interval of the integral become

#x in [0,2]#

now let set up the integral of Area

#color(red)[A=int_a^by_2-y_1*dx#

#A=int_0^2(4x-x^2)-(x^2)*dx#

#A=int_0^(2)4x-2x^2*dx=[2x^2-2/3x^3]_0^2#

#[8-16/3]=(24-16)/3=8/3=2.667 (unite)^2#

show the graph of the area below:

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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