91Ó°ÊÓ

The Fundamental Theorem of Calculus is a crucial link between differentiation and integration. It tells us that if a function is continuous on an interval, then it can be integrated, and the integral can give back the original function through differentiation. This theorem is divided into two parts.

• The **first part** states that if you have a continuous function and you integrate it over an interval, the antiderivative of that function at the upper bound minus the antiderivative at the lower bound will give you the integral.
• The **second part** involves differentiation and states that if you take the derivative of an integral of a function, you get back that function. Essentially, it says that if you have a function that is the integral of another function, differentiating it basically "undoes" the integration process, leaving you with the original function.

For instance, in our problem, the function \( F(x) = \int_{0}^{x^2} \sqrt{1-t^2} dt \) uses this concept. By applying the Fundamental Theorem of Calculus, we can differentiate under the integral symbol and find \( F'(x) \).

The Chain Rule is a key concept in calculus for understanding how to differentiate composite functions. It's a method for finding the derivative of a function that is composed of two or more functions. Simply put, if you have a function within a function, the Chain Rule helps you find the derivative.

• When using the Chain Rule, you multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function.
• In terms of formula, if you have a composition \( f(g(x)) \), the derivative \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).

In the original problem, the Chain Rule is used in conjunction with the Fundamental Theorem of Calculus to find \( F'(x) \). We differentiate \( \int_{0}^{x^2} \sqrt{1-t^2} dt \) to get \( 2x\sqrt{1-x^4} \). The \( 2x \) comes from the derivative of the inner function \( x^2 \).

The derivative of an integral is a special application of the Fundamental Theorem of Calculus. It involves functions defined by definite integrals with variable limits. This method helps in evaluating how the accumulated value of a function changes as its integration bounds vary.
For example, if \( F(x) = \int_{a}^{u(x)} f(t) dt \), then by the Fundamental Theorem of Calculus Part 2, the derivative \( F'(x) = f(u(x)) \cdot u'(x) \). Here, \( f(u(x)) \) is evaluated at the upper limit function \( u(x) \), and \( u'(x) \) is the derivative of the upper limit function with respect to \( x \).
Applying this to our exercise, we have the integral from \( 0 \) to \( x^2 \) of \( \sqrt{1-t^2} \). Using this concept, we differentiate to find that \( F'(x) = 2x\sqrt{1-x^4} \), where \( \sqrt{1-x^4} \) is \( f(x^2) \) and \( 2x \) is the derivative of \( x^2 \).

Understanding the domain of a function is essential in calculus. The domain is the set of all possible input values (typically \( x \) values) for which the function is defined. Restrictions on domain often come from domains of square roots, logarithms, denominators of fractions, etc.

• For polynomial functions, the domain is usually all real numbers.
• For square root functions, the expression inside the square root must be non-negative.

In the given problem, the function \( \sqrt{1-t^2} \) is under a square root and hence must be non-negative. This restriction gives \(-1 \leq t \leq 1\). Since \( F(x) = \int_{0}^{x^2} \sqrt{1-t^2} dt \), we must ensure \( 0 \leq x^2 \leq 1 \), leading us to conclude that the domain of \( F(x) \) is \( [0, 1] \). This reflects the need to ensure that the integral's bounds are meaningful and valid within the function's calculations.