Find the area between the graph of f and
the x-axis over the interval [−4, 1] when
f(x) = 4 + 4x.
1. Area = 34 sq. units
2. Area = 30 sq. units
3. Area = 28 sq. units
4. Area = 32 sq. units
5. Area = 26 sq. units
(5)
You just have two triangles. Doing a single integral uses signed areas, so you'd end up with the wrong answer.
To find the area between the graph of f(x) = 4 + 4x and the x-axis over the interval [-4, 1], we can use the definite integral.
The definite integral of a function f(x) over an interval [a, b] gives the signed area between the graph of the function and the x-axis over that interval. To find the area, we need to evaluate the integral of the absolute value of the function.
In this case, f(x) = 4 + 4x, and the interval is [-4, 1]. So, we need to evaluate the integral of the absolute value of f(x) from -4 to 1:
∫[from -4 to 1] |f(x)| dx
First, let's find the absolute value of f(x). Since f(x) = 4 + 4x, the absolute value of f(x) is equal to |4 + 4x|.
Next, we need to evaluate the integral of |f(x)| from -4 to 1. This can be done by splitting the interval into two parts: [-4, 0] and [0, 1].
For the interval [-4, 0], we have:
∫[from -4 to 0] (4 + 4x) dx
Integrating this expression gives us:
∫[from -4 to 0] (4x + 4x^2/2) dx
= [2x^2 + 4x^3/3] from -4 to 0
= (2(0)^2 + 4(0)^3/3) - (2(-4)^2 + 4(-4)^3/3)
= (0 + 0) - (2(16) + 4(-64)/3)
= -32 - (-256/3)
= -32 + 256/3
= 160/3
For the interval [0, 1], we have:
∫[from 0 to 1] (4 + 4x) dx
Integrating this expression gives us:
∫[from 0 to 1] (4x + 4x^2/2) dx
= [2x^2 + 4x^3/3] from 0 to 1
= (2(1)^2 + 4(1)^3/3) - (2(0)^2 + 4(0)^3/3)
= (2 + 4/3) - (0 + 0)
= 2 + 4/3
= 10/3
To find the total area between the graph of f(x) and the x-axis, we sum up the areas over both intervals:
Total area = (160/3) + (10/3)
= 170/3
Therefore, the correct answer is not among the given options. The area between the graph of f(x) = 4 + 4x and the x-axis over the interval [-4, 1] is 170/3 square units.