Find the area enclosed by curves y=4x^2 and y=x+3 ?

1 Answer
Jun 24, 2018

#color(blue)("Area"=343/96 " units"^2#

Explanation:

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First we need to find the points of intersection of #f(x)=4x^2# and g(x)=x+3#

#:.#

#4x^2=x+3#

#4x^2-x-3=0#

#(4x+3)(x-1)=0=>x=-3/4 and x=#

These will be our lower and upper bounds of integration.

By observing the graphs above, we can see that, if we find the area under #g(x)# between the the bounds we just calculated and then subtract the area #f(x)# between these bounds, then we are left with the area between the curves. Since integration is distributive over the sum we can simplify things by subtracting #f(x)# from #g(x)# first.

#x+3-4x^2#

Integrate this:

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

You can ignore the constant #k# this vanishes in calculating area.

Using the conventional method, we get:

#"Area"=int_(-3/4)^(1)(x+3-4x^2)dx=[-4/3x^3+1/2x^2+3x]_(-3/4)^(1)#

#=[-4/3x^3+1/2x^2+3x]^(1)-[-4/3x^3+1/2x^2+3x]_(-3/4)#

Plugging in the upper and lower bounds:

#=[-4/3(1)^3+1/2(1)^2+3(1)]^(1)#

#-[-4/3(-3/4)^3+1/2(-3/4)^2+3(-3/4)]_(-3/4)#

#"Area"=[13/6]-[-45/32]=343/96 " units"^2#

Final area plot:

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