Can somebody explain how to get area between f(x) = x²-4x & g(x)=0 [Urgent]

[ricecrispie]

Can somebody explain how to get area between f(x) = x²-4x & g(x)=0 [Urgent]

Hi all ! I have a test in a couple hours and I'm stuck on this question:

Find the region bounded by the graphs of the functions and find the area:

f(x) = x²-4x
g(x)=0

So the Area is the parabola under the x-axis:

points of intersection:
x(x-4) = 0
x = 0 or = 4

A = ∫ x²-4x dx
= x³/3 - 2x² |⁴₀ = - 32/3

But the answer should be positive, since the g(x) is above f(x) should my integral read 0-(x²-4x) dx ??

I want to understand the logic behind this as I am sure it will come up in more complex questions. Any feedback is appreciated

Sent from my LG-H840 using Tapatalk

 

[Dr.Peterson]

Hi all ! I have a test in a couple hours and I'm stuck on this question:

Find the region bounded by the graphs of the functions and find the area:

f(x) = x²-4x
g(x)=0

So the Area is the parabola under the x-axis:

points of intersection:
x(x-4) = 0
x = 0 or = 4

A = ∫ x²-4x dx
= x³/3 - 2x² |⁴₀ = - 32/3

But the answer should be positive, since the g(x) is above f(x) should my integral read 0-(x²-4x) dx ??

I want to understand the logic behind this as I am sure it will come up in more complex questions. Any feedback is appreciated

Yes, you should be integrating the top function minus the bottom function, since that gives the height of the element of area.

 

[Noismaker]

Hi all ! I have a test in a couple hours and I'm stuck on this question:

Find the region bounded by the graphs of the functions and find the area:

f(x) = x²-4x
g(x)=0

So the Area is the parabola under the x-axis:

points of intersection:
x(x-4) = 0
x = 0 or = 4

A = ∫ x²-4x dx
= x³/3 - 2x² |⁴₀ = - 32/3

But the answer should be positive, since the g(x) is above f(x) should my integral read 0-(x²-4x) dx ??

I want to understand the logic behind this as I am sure it will come up in more complex questions. Any feedback is appreciated

Sent from my LG-H840 using Tapatalk