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.
F(x) = ∫(2t + 4) dt from 2 to x
= t^2 + 4t | from 2 to x
= x^2 + 4x - (4 + 2*4)
= x^2 + 4x - (4 + 8)
=> x^2 + x - 12
Yes the answer is x^2 + x - 12
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To write the function F(x) using the given graph of f(t) = 2t + 4 on the interval [−4, 6], we need to find the integral of f(t) and define F(x) as the integral from 2 to x of f(t) dt.
1. First, let's find the integral of f(t) = 2t + 4. To integrate, we add 1 to the exponent of t and divide by the new exponent. The integral of 2t + 4 becomes t^2 + 4t.
2. Next, we will replace the variable t with x in the integral expression to get F(x). Since the integral is taken from 2 to x, we have F(x) = ∫[2,x] (t^2 + 4t) dt.
3. Now, we can evaluate the definite integral from 2 to x. To do this, we substitute the upper limit (x) and the lower limit (2) into the antiderivative expression and subtract the result at the lower limit from the result at the upper limit: F(x) = [t^2 + 4t] evaluated from 2 to x.
F(x) = ([x^2 + 4x] - [2^2 + 4*2])
4. Simplify the expression: F(x) = (x^2 + 4x - 8).
Therefore, the function F(x) based on the given graph is F(x) = (x^2 + 4x - 8).
F(x) = ∫(2t + 4) dt from 2 to x
= t^2 + t | from 2 to x
= x^2 + x - (4 + 2)
= x^2 + x - 6