F(x) = ∫[1,x] f(t) dt
= ∫[1,x] 2t+2 dt
= t^2+2t [1,x]
= (x^2+2x)-(1^2+2*1)
= x^2+2x-3
F(x) = x2 + 3x
F(x) = x2 + 2x – 12
F(x) = x2 + 2x – 3
F(x) = x2 + 4x – 8
= ∫[1,x] 2t+2 dt
= t^2+2t [1,x]
= (x^2+2x)-(1^2+2*1)
= x^2+2x-3
Since f(x) equals the integral from 1 to x of f(t) dt, we need to find the area under the curve of f(t) from 1 to x. Looking at the graph of f(t) = 2t + 2, we can see that it's a straight line with a slope of 2 and a y-intercept of 2.
To find F(x), we need to integrate f(t) from 1 to x. The integral of 2t + 2 with respect to t is t^2 + 2t. Now let's substitute the limits of integration:
F(x) = x^2 + 2x - (1^2 + 2(1))
Simplifying further:
F(x) = x^2 + 2x - 3
So, the correct answer is F(x) = x^2 + 2x - 3.-Ta-da!
Step 1: Find the integral of f(t) = 2t + 2.
∫(2t + 2) dt = t^2 + 2t + C
Step 2: Evaluate the integral from 1 to x.
F(x) = ∫(2t + 2) dt from 1 to x
= [t^2 + 2t] from 1 to x
= (x^2 + 2x) - (1^2 + 2(1))
= x^2 + 2x - 3
Therefore, the function F(x) is F(x) = x^2 + 2x - 3.
Given that f(t) = 2t + 2, we can integrate f(t) to find F(x):
∫(2t + 2) dt = t^2 + 2t + C,
where C is the constant of integration.
Now, we want to evaluate F(x) from 1 to x:
F(x) = [t^2 + 2t + C] evaluated from 1 to x
To find F(x), we substitute x into the expression t^2 + 2t + C and subtract the value at t = 1:
F(x) = (x^2 + 2x + C) - (1^2 + 2(1) + C)
Simplifying,
F(x) = (x^2 + 2x + C) - (1 + 2 + C)
F(x) = x^2 + 2x + C - 3
Since C is a constant and we don't have its specific value, we can express it as another constant. Let's denote it as C' for simplicity:
F(x) = x^2 + 2x + C' - 3
Thus, the function F(x) is represented by the equation:
F(x) = x^2 + 2x + C' - 3.
Therefore, none of the given options, F(x) = x^2 + 3x, F(x) = x^2 + 2x – 12, F(x) = x^2 + 2x – 3, and F(x) = x^2 + 4x – 8, are correct.