Find the area of the region between the parabola 4y = x^2 and the line 3x + 4 y = 4
The curves intersect at (-4,4) and (1,1/4)
So, just integrate to find
a = ∫[-4,1] (4-3x)/4 - x^2/4 dx
a = 125/24
To find the area between the parabola 4y = x^2 and the line 3x + 4y = 4, we can start by finding the points of intersection between the two curves. Then, we can integrate to find the area under the curves.
Let's solve the two equations to find the points of intersection:
1. Parabola: 4y = x^2
Rearrange the equation to find y in terms of x:
y = x^2/4
2. Line: 3x + 4y = 4
Rearrange the equation to find y in terms of x:
4y = 4 - 3x
y = (4 - 3x)/4
Now, we can set the two equations equal to each other to find the x-values of the points of intersection:
x^2/4 = (4 - 3x)/4
Multiply both sides by 4 to get rid of the fractions:
x^2 = 4 - 3x
Rearrange the equation:
x^2 + 3x - 4 = 0
This is a quadratic equation. We can factor it or use the quadratic formula to find the values of x:
(x + 4)(x - 1) = 0
So we have two possible x-values: x = -4 and x = 1.
Now that we have the x-values, we can substitute them into either equation to find the y-values of the points of intersection:
For x = -4:
y = (-4)^2/4 = 4
For x = 1:
y = (1)^2/4 = 1/4
Therefore, the two points of intersection are (-4, 4) and (1, 1/4).
Now, to find the area between the curves, we will integrate the difference between the two functions:
∫[a, b] (f(x) - g(x)) dx
Where f(x) is the upper curve (parabola) and g(x) is the lower curve (line).
In this case, we need to integrate the function (x^2/4 - (4 - 3x)/4) from -4 to 1, since these are the x-values of the points of intersection:
Area = ∫[-4, 1] (x^2/4 - (4 - 3x)/4) dx
Evaluating this integral will give you the area between the parabola and the line.