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?
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!
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
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