Questions: Use the Fundamental Theorem of Calculus to find an expression for the derivative of the given function defined on the given interval, if it exists. F(x) = integral from 1 to x of 3 dt, [1,1000] F'(x) = 2 on [1,1000] F'(x) = 3 on [1,1000] F'(x) = 0 on [1,1000] F'(x) is not defined over all of [1,1000]. None of these

Use the Fundamental Theorem of Calculus to find an expression for the derivative of the given function defined on the given interval, if it exists.

F(x) = integral from 1 to x of 3 dt, [1,1000]

F'(x) = 2 on [1,1000]

F'(x) = 3 on [1,1000]

F'(x) = 0 on [1,1000]

F'(x) is not defined over all of [1,1000].

None of these

Transcript text: Use the Fundamental Theorem of Calculus to find an expression for the derivative of the given function defined on the given interval, if it exists. \[ F(x)=\int_{1}^{x} 3 d t, \quad[1,1000] \] $F^{\prime}(x)=2$ on $[1,1000]$ $F^{\prime}(x)=3$ on $[1,1000]$ $F^{\prime}(x)=0$ on $[1,1000]$ $F^{\prime}(x)$ is not defined over all of $[1,1000]$. None of these

Solution

Solution Steps

To find the derivative of the function $ F(x) = \int_{1}^{x} 3 \, dt $, we can apply the Fundamental Theorem of Calculus. This theorem states that if $ F(x) = \int_{a}^{x} f(t) \, dt $, then $ F'(x) = f(x) $. In this case, the integrand $ f(t) = 3 $ is a constant function. Therefore, the derivative $ F'(x) $ is simply the constant function $ f(x) = 3 $.

Step 1: Define the Function

We are given the function defined by the integral: \[ F(x) = \int_{1}^{x} 3 \, dt \]

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if $ F(x) = \int_{a}^{x} f(t) \, dt $, then the derivative $ F'(x) $ is given by: \[ F'(x) = f(x) \] In this case, since $ f(t) = 3 $ is a constant function, we have: \[ F'(x) = 3 \]

Step 3: Conclusion

The derivative $ F'(x) $ exists and is equal to 3 for all $ x $ in the interval $[1, 1000]$.