Let f be the function defined by
f(x) = ∫[0 to x] (4t^3 + 2t - 1) dt.
(a) Find f'(x).
(b) Determine the value of f(1) using the Fundamental Theorem of Calculus.
(a) To find f'(x), we can use the Fundamental Theorem of Calculus, which states that if a function f(x) is defined as the integral of another function F(x) with respect to x, then the derivative of f(x) is equal to F(x).
Using this theorem, we can find f'(x) by finding the derivative of the function inside the integral:
f(x) = ∫[0 to x] (4t^3 + 2t - 1) dt.
To find the derivative, we apply the Power Rule and the Constant Rule:
f'(x) = d/dx [∫[0 to x] (4t^3 + 2t - 1) dt] = (4x^3 + 2x - 1).
Therefore, the derivative of f(x) is f'(x) = 4x^3 + 2x - 1.
(b) To determine the value of f(1) using the Fundamental Theorem of Calculus, we substitute x = 1 into the original function f(x) = ∫[0 to x] (4t^3 + 2t - 1) dt:
f(1) = ∫[0 to 1] (4t^3 + 2t - 1) dt.
To evaluate this definite integral, we find the antiderivative of the integrand:
F(t) = ∫ (4t^3 + 2t - 1) dt = t^4 + t^2 - t + C,
where C is the constant of integration.
Now, we can evaluate the definite integral from 0 to 1:
f(1) = F(1) - F(0) = (1)^4 + (1)^2 - (1) - [(0)^4 + (0)^2 - (0)] = 1 + 1 - 1 - 0 = 1.
Therefore, using the Fundamental Theorem of Calculus, we find that f(1) = 1.
(Note: The actual value of C is not necessary for this problem, as it cancels out when evaluating the definite integral.)