Find the area bounded by the following functions y=4x-x2 and y=3
here's where you went wrong:
= (18 - 9 - 9) - (2 - 1/3 - 3)
= 0 - (1/3) - 1
= -4/3
It should have been
= (18 - 9 - 9) - (2 - 1/3 - 3)
= 0 - (-1/3 - 1)
= 4/3
The graphs clearly show you should expect a positive algebraic area
Apologies for the mistake. Thank you for catching that error. The correct calculation should be as follows:
= (18 - 9 - 9) - (2 - 1/3 - 3)
= 0 - (-1/3 - 1)
= 4/3
Therefore, the correct area bounded by the functions y=4x-x^2 and y=3 is 4/3 square units. Thank you for pointing out the error.
To find the area bounded by the two functions y=4x-x^2 and y=3, we first need to find the points where the two functions intersect.
Setting the two functions equal to each other, we get:
4x-x^2 = 3
Rearranging terms, we get:
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
So, x = 1 or x = 3
Now we can find the area by integrating the difference between the two functions from x=1 to x=3:
Area = ∫[1,3] (4x - x^2 - 3) dx
= [2x^2 - (x^3)/3 - 3x] [1,3]
= (18 - 9 - 9) - (2 - 1/3 - 3)
= 0 - (1/3) - 1
= -4/3
Thus, the area bounded by the two functions y=4x-x^2 and y=3 is 4/3 square units.