Use the graph of f(t) = 2t − 5 on the interval [−2, 10] to write the function F(x), where f of x equals the integral from 1 to x of f of t dt.
Answer choices:
F(x) = 2x − 5
F(x) = x2 − 5x − 4
F(x) = x2 − 5x + 4
F(x) = x2 − 5x + 36
F(x) =
x
∫ ( 2 t - 5 ) dt =
1
x
( 2 t² / 2 - 5 t ) =
1
x
( t² - 5 t ) =
1
x² - 5 x - ( 1² - 5 ∙ 1 ) = x² - 5 x - ( 1 - 5 ) = x² - 5 x - ( - 4 ) = x² - 5 x + 4
nicely done, but it's clear that the graph was not necessary.
Well, let's see here. We want to find the function F(x), which represents the integral from 1 to x of f(t) dt. To do that, we need to find the antiderivative of f(t). The function f(t) = 2t - 5 is a linear function, so its antiderivative will be a quadratic function.
If we integrate 2t, we get t^2, and if we integrate -5, we get -5t. So, the antiderivative of f(t) is F(t) = t^2 - 5t.
To find F(x), we need to evaluate F(t) at the upper limit of integration, x, and subtract F(t) evaluated at the lower limit of integration, 1. So, F(x) = F(x) - F(1).
Let's calculate F(x):
F(x) = (x^2 - 5x) - (1^2 - 5*1)
= x^2 - 5x - 1 + 5
= x^2 - 5x + 4.
So, the function F(x) is F(x) = x^2 - 5x + 4. That's the third option!
To write the function F(x) using the graph of f(t) = 2t − 5 on the interval [−2, 10], we need to evaluate the integral of f(t) with respect to t from 1 to x.
First, let's find the definite integral of f(t) = 2t − 5 from 1 to x:
∫[1, x] (2t − 5) dt
Using the integrative properties, we can integrate each term separately:
∫[1, x] 2t dt - ∫[1, x] 5 dt
The integral of 2t with respect to t is t^2, and the integral of a constant (like 5) with respect to t is that constant multiplied by t:
t^2 - 5t [integrating the terms]
Now, let's evaluate this expression by plugging in the limits of integration [1, x]:
F(x) = t^2 - 5t │[1, x]
Substituting x into the expression:
F(x) = x^2 - 5x - (1^2 - 5(1))
= x^2 - 5x - 1 + 5
= x^2 - 5x + 4
Therefore, the function F(x) is equal to x^2 - 5x + 4.
To find the function F(x), we need to integrate the function f(t) from 1 to x.
First, let's find the integral of f(t) = 2t - 5 with respect to t. The integral of 2t with respect to t is t^2, and the integral of -5 with respect to t is -5t. Therefore, the integral of f(t) is given by:
∫f(t)dt = ∫(2t - 5)dt = t^2 - 5t
Now, we need to evaluate this integral from 1 to x:
F(x) = ∫[1, x] f(t)dt
To do this, we substitute the upper limit (x) into the integral expression:
F(x) = ∫[1, x] (2t - 5) dt
Now, we integrate the function (2t - 5):
F(x) = [t^2 - 5t] evaluated from 1 to x
Substituting the upper limit (x) into the expression:
F(x) = (x^2 - 5x) - (1^2 - 5*1)
Simplifying:
F(x) = x^2 - 5x - 1 + 5
F(x) = x^2 - 5x + 4
Therefore, the function F(x) is F(x) = x^2 - 5x + 4.
The correct answer is: F(x) = x^2 - 5x + 4.