Use the graph of f(t) = 2t + 2 on the interval [-1, 4] to write the function F(x), where f(x) = x∫-1 f(t) dt
a. F(x) = x^2 + 3x
b. F(x) = x^2 + 2x - 12 (my answer)
c. F(x) = x^2 + 2x - 3
d. F(x) = x^2 + 4x - 8
sorry the integral denotation is x∫1 (not x∫-1)
integral is t^2 + 2 t + constant
so at t = x minus at t = 1
x^2 + 2 x + c - 1^2 -2 - c
= x^2 + 2 x - 3
Thank you!
To find the function F(x) given f(t) = 2t + 2 on the interval [-1, 4], you need to evaluate the definite integral of f(t) with respect to t from -1 to x.
The integral of f(t) with respect to t gives the area under the curve f(t) between the limits of integration. Since f(t) = 2t + 2 is a linear function, the area can be represented as the area of a trapezoid.
To find the area under the curve f(t) between -1 and x, we can compute the integral:
∫(-1 to x) (2t + 2) dt
To evaluate this integral, we need to find the antiderivative of (2t + 2). Integrating (2t + 2) with respect to t gives t^2 + 2t + C, where C is the constant of integration.
Now, we can substitute the limits of integration:
F(x) = ∫(-1 to x) (2t + 2) dt
= [t^2 + 2t] from -1 to x
= (x^2 + 2x) - (-1^2 + 2(-1))
= (x^2 + 2x) - (1 - 2)
= (x^2 + 2x) + 1 - 2
= x^2 + 2x - 1
Therefore, the correct answer is c. F(x) = x^2 + 2x - 3.