Find the area of the region bounded by y=4-x^2 & y=2x+1 =?

1 Answer
Mar 21, 2018

#32/3# cubic units

Explanation:

The area we seek lies between the two curves. First we need to find the points of intersection between these:

#4-x^2=2x+1#

#x^2+2x-3=0#

#(x-1)(x+3)=0=>x=-3 and x=1#

Interval #[-3,1]#

Testing a value in this interval with our two functions, say #x=0#

#4-(0)^2=4#

#2(0)+1=1#

So, #4-x^2# attains greater values and is therefore above #2x+1# in this interval.

We can therefore find the area under #4-x^2# and then subtract the area under #2x+1# from this. i.e.

Our lower and upper bounds are #-3 ,1#

Integrals:

#int_(-3)^(1)(4-x^2)dx-int_(-3)^(1)(2x+1)dx#

Since the integral is distributive over the sum:

#int_(-3)^(1)[(4-x^2)-(2x+1)]dx#

Simplifying:

#int_(-3)^(1)[(3-2x-x^2]dx#

A#=int_(-3)^(1)[(3-2x-x^2]dx =[3x-x^2-1/3x^3]_(-3)^(1)#

A#=[3x-x^2-1/3x^3]^(1)-[3x-x^2-1/3x^3]_(-3)#

Plugging in upper and lower bounds:

#A=[3(1)-(1)^2-1/3(1)^3]^(1)-[3(-3)-(-3)^2-1/3(-3)^3]_(-3)#

#A=[5/3]-[-27/3]=32/3 \ \ \ \ # cubic units.

GRAPH:

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