The Fundamental Theorem of Calculus provides a powerful connection between differentiation and integration. In this case, we want to find the derivative of the function defined by the integral of a function with respect to a variable. Let’s break down the steps required to use Part I of the theorem effectively.
1. **Understand the Integral**: First, we need to understand what the function f(x) is defined as. In your question, it seems you meant to express it as:
f(x) = ∫ (31 + 3t^2 + 17) dt from a to x
For clarity, let’s assume that ‘a’ is a constant limit and ‘x’ is the upper limit of integration. The expression in the integral is simply a polynomial.
2. **Apply Part I of the Fundamental Theorem of Calculus**: According to the theorem, if you have a continuous function g(t) and f(x) is defined as the integral of g(t) from a to x, i.e.,
f(x) = ∫ (g(t)) dt from a to x,
then the derivative of f(x) with respect to x is the function g evaluated at x:
f'(x) = g(x).
3. **Find g(t)**: In our case, the function g(t) = 31 + 3t^2 + 17 simplifies to:
g(t) = 48 + 3t^2.
4. **Calculate the Derivative**: To find f'(x), we will substitute x into the function g(t):
f'(x) = g(x) = 48 + 3(x^2).
5. **Conclusion**: Thus, the derivative of the function f(x) defined by the integral is:
f'(x) = 48 + 3x^2.
This process clearly illustrates the usage of Part I of the Fundamental Theorem of Calculus to transition from an integral to its derivative, showcasing the relationship between these two fundamental concepts in calculus.