Use the graph of f(t) = 2t + 4 on the interval [−4, 6] to write the function F(x), where f of x equals the integral from 2 to x of f of t dt.
To write the function F(x), where f(x) equals the integral from 2 to x of f(t) dt, we need to calculate the definite integral of f(t) from 2 to x.
Let's start by understanding the graph of f(t) = 2t + 4 on the interval [-4, 6].
The graph of f(t) = 2t + 4 is a straight line with a slope of 2 and a y-intercept of 4. On the given interval [-4, 6], the graph starts at (-4, -4) and ends at (6, 16).
To find F(x), we need to calculate the integral of f(t) from 2 to x. To do this, we can find the area between the graph of f(t) and the x-axis on the interval [2, x].
First, let's determine the equation for the graph of f(t) on the interval [2, x]. Since f(t) = 2t + 4, we substitute t with x:
f(x) = 2x + 4
Now, we can find the definite integral of f(x) from 2 to x:
F(x) = ∫[2, x] f(t) dt
F(x) = ∫[2, x] (2t + 4) dt
To calculate the integral, we need to apply the rules of integration. Integrating 2t gives us t^2, and integrating 4 gives us 4t.
F(x) = [t^2 + 4t] evaluated from 2 to x
F(x) = (x^2 + 4x) - (2^2 + 4*2)
F(x) = (x^2 + 4x) - (4 + 8)
F(x) = x^2 + 4x - 12
Therefore, the function F(x) is given by F(x) = x^2 + 4x - 12.