Explain the Fundamental Theorem of Calculus Part 1 (the antiderivative part) and the Fundamental Theorem of Calculus Part 2 (the evaluation part). Then, compare each part and specify when one part would be used over the other. Use relevant key words in your writing. Give a real-world example that involves Part 2 of the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus Part 1 explains the relationship between differentiation and integration, whereas the Fundamental Theorem of Calculus Part 2 explains how to evaluate a definite integral.

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a) The Fundamental Theorem of Calculus Part 1 (antiderivative part)

This theorem explains the relationship between differentiation and integration. It states the following:

If `f` is a continuous function over an interval `[a, b]` and the function `F(x)`is given by `F(x) = \int_{a}^{x} f(y) dy` , then `F(x)` is continuous and differentiable on `[a, b]` , and the derivative of `F(x)`

is `f(x)`, i.e., `F^{'}(x) = f(x)` .

Thus, this theorem ensures that every function that is continuous in a given interval has an antiderivative.

b) The Fundamental Theorem of Calculus Part 2 (the evaluation part)

This theorem states the following:

If `f` is a continuous function over the interval `[a, b]` and `F(x)` is any one of the antiderivatives of `f(x)` , then `\int_{a}^{b} f(x)dx` = `F(b) - F(a)` .

In other words, this theorem states that if there exists an antiderivative for `f(x)` , then the definite integral `\int_{a}^{b} f(x) dx` can be computed by evaluating this antiderivative at the endpoints of the stated interval.

c) The first part of the theorem (the antiderivative part) is used to define antiderivatives using definite integrals, whereas the second part of the theorem (evaluation part) is used to evaluate definite integrals.

d) Evaluating an example using the Fundamental Theorem of Calculus Part 2

Evaluate the integral `\int_{0}^{2} 4x^{3} - 3x^{2} dx`

Using the Fundamental Theorem of Calculus Part 2, we obtain the following:

`[\frac{4x^{4}}{4} - \frac{3x^{3}}{3} ]_{0}^{2}` = `[x^{4} - x^{3}]_{0}^{2} = (16 - 8) - 0 = 8`

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