How do you find the area between the curves #y=4x-x^2# and #y=x#?

2 Answers
Feb 20, 2015

First, we need to find where the two functions intersect.

#4x-x^2=x #

#x^2-4x+x=0 #

#x^2-3x=0 #

#x(x-3)=0 #

#x=0 # and #x=3 #

So our interval of integration is

#0<=x<=3 #

#y=4x-x^2 # is a parabola that is concave down

#y=x# is the line that passes through the origin with slope 1

The integral for the area is

#int4x-x^2-xdx=int3x-x^2dx #

integrating we have

#3/2x^2-1/3x^3 #

Evaluating at we have

#(3/2)3^2-(1/3)3^3-0=27/2-27/3=27/2-9 #

#81/6-54/6=27/6=9/2 #

Feb 20, 2015

Find the points of intersection, #(a,y_a)# and #(b,y_b)#
between
#y = 4x - x^2# and #y = x#
then subtract from the integral of the first (between #a# and #b#)
the integral of the second (again, between #a# and #b#)

Part 1:
Points of intersection occurs when
#4x - x^2 = x#
This occurs when either #x = 0# or #x = 3#
(we could, but don't actually need to calculate #y_a# and #y_b#)

Part 2:
#int_a ^b (4x-x^2) dx - int_a^b x dx#

# = ((2x^2 - x^3/3) |_a^b) - (((x^2)/2) |_a^b)#

with #a = 0# and #b = 3#
this becomes
# ((2*9 - 9/3) - (0)) - ( (9/2) - (0) )#
#= 10.5#

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