Finding area of the bounded region
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Finding area of the bounded region

[From: ] [author: ] [Date: 11-12-15] [Hit: ]
x^2 - 4x - 5 = x^3 - 3x^2 - 5 ==> x = 0 and x = 2,so we will be integrating over [0, 2].To determine which function is higher, we can choose any value of x in (0, 2) and plug it into both functions.......
f(x)=x^2 - 4x - 5
g(x) = x^3 - 3x^2 - 5

Find the area A of the bounded region determined by the graphs of f and g


How do you know which function is upper and lower without graphing with a calculator?
I tried draw a graph on a paper but it's not going well..
Can someone please go through this?

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By equating f(x) and g(x), we see that they intersect when:
x^2 - 4x - 5 = x^3 - 3x^2 - 5 ==> x = 0 and x = 2,

so we will be integrating over [0, 2].

To determine which function is higher, we can choose any value of x in (0, 2) and plug it into both functions. Whatever function gives the highest value will be the top function and the other will be the bottom function.

When x = 1, we see that:
f(1) = 1 - 4 - 5 = -8 and g(1) = 1 - 3 - 5 = -7 ==> g(1) > f(1).

Thus, g(x) is the upper function and f(x) is the lower. Therefore, the required area is:
A = ∫ [(x^3 - 3x^2 - 5) - (x^2 - 4x - 5)] dx (from x=0 to 2)
= ∫ (x^3 - 4x^2 + 4x) dx (from x=0 to 2)
= [(1/4)x^4 - (4/3)x^3 + 2x^2] (evaluated from x=0 to 2)
= (4 - 32/3 + 8) - (0 - 0 + 0)
= 4/3 units^2.

I hope this helps!

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x^2 -4x - 5 = x^3 - 3x^2 - 5

5 cancel out move it all to one side

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

with grouping can be written
(x)(2-x)(-2+x) = 0

x = 0 x = 2

0 is lower 2 is upper for the range

at f(1) = -8 g(1) = -7

so its going to be g(x)-f(x)

So you will have the integration of g(x)-f(x) from a=0 t b = 2
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