![Create a visually appealing graphic representing a mathematical concept. It should depict the graph of a function, specifically f(t) = 2t + 2, within the interval [–1, 4]. In parallel, it should visualize the concept of an integral from 1 to x of this function f(t), giving an insight into the process of integral calculus. The image should be illustrative and easy to understand, make sure to use colors and shapes to differentiate the various elements of the graph. However, it should not include any text or numbers.](/images/questions/1677724/1024x1024.png)
Use the graph of f(t) = 2t + 2 on the interval [–1, 4] to write the function F(x), where f of x equals the integral from 1 to x of f of t dt.
F(x) = x2 + 3x
F(x) = x2 + 2x – 12
F(x) = x2 + 2x – 3
F(x) = x2 + 4x – 8
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
Thank you
Well, I must say this question really put my mathematical skills to the test! But fear not, the Clown Bot is here to clown around and hopefully bring a smile to your face. Now, let's calculate F(x) using the given graph.
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!
To find the function F(x) using the graph of f(t) = 2t + 2 on the interval [–1, 4], we need to evaluate the integral of f(t) from 1 to x.
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.
To write the function F(x), we need to find the integral of f(t) with respect to t and evaluate it from 1 to x.
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.