Find the area of the region bounded by the graphs of y = −x2 + 3x + 4 and
y = 4.
a) 2.7
b) 4.5
c) 28.5
d) 16.5
To find the area of the region bounded by the two graphs, we need to find the points where they intersect.
First, let's set the two equations equal to each other:
4 = -x^2 + 3x + 4
Now, let's rearrange the equation to get it in standard form:
x^2 - 3x = 0
Next, let's factor out an x from the left side:
x(x - 3) = 0
This equation tells us that x = 0 or x = 3.
Now, let's substitute these values of x into either of the original equations to find the corresponding y-values.
When x = 0:
y = -0^2 + 3(0) + 4 = 4
When x = 3:
y = -(3^2) + 3(3) + 4 = -9 + 9 + 4 = 4
So, the two graphs intersect at (0, 4) and (3, 4).
To find the area between the two graphs, we need to integrate the difference between the equations with respect to x over the interval [0, 3].
Using the formula for finding the area between two curves, the formula becomes:
Area = ∫[0, 3] (-x^2 + 3x + 4) - 4 dx
Now, let's integrate this expression:
Area = ∫[0, 3] (-x^2 + 3x + 4 - 4) dx
= ∫[0, 3] (-x^2 + 3x) dx
= [-x^3/3 + 3x^2/2] | [0, 3]
= [-(3^3)/3 + 3(3)^2/2] - [-(0^3)/3 + 3(0)^2/2]
= [-27/3 + 27/2] - [0]
= [-9 + 13.5] - [0]
= 4.5
Therefore, the area of the region bounded by the graphs is 4.5.
The answer is (b) 4.5.
the graphs intersect at x = 0,3 so the area is just
∫[0,3] (-x^2+3x+4) - 3 dx