Problem 57 - Calculus Single Variable

Question

\(F(x)\) is a function of a variable \(x\) that appears in a limit (or in the limits) of integration of a given definite integral. Express \(F(x)\) explicitly by calculating the integral. $$ F(x)=\int_{1}^{x^{2}} \frac{1}{\sqrt{t}} d t $$

Step-by-Step Solution

Answer

The function \( F(x) \) is \( 2x - 2 \).

Step 1: Recognize the Integral Form

The given function \( F(x) = \int_{1}^{x^2} \frac{1}{\sqrt{t}} \, dt \) is a definite integral with variable upper limit. We need to integrate the function \( \frac{1}{\sqrt{t}} \) with respect to \( t \).

Step 2: Find the Indefinite Integral

First, find the indefinite integral of \( \frac{1}{\sqrt{t}} \). The antiderivative of \( \frac{1}{\sqrt{t}} \) is \( 2\sqrt{t} \), since the derivative of \( 2\sqrt{t} \) is \( \frac{1}{\sqrt{t}} \).

Step 3: Compute the Definite Integral

Calculate the definite integral by evaluating the antiderivative at the upper and lower limits. Substituting, we get:\[F(x) = \left[ 2\sqrt{t} \right]_1^{x^2} = 2\sqrt{x^2} - 2\sqrt{1}\]

Step 4: Simplify the Expression

Simplify the expression obtained from evaluating the definite integral. We have:\[2\sqrt{x^2} = 2x \quad\text{and}\quad 2\sqrt{1} = 2\]Therefore, the function becomes \( F(x) = 2x - 2 \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus, connecting differentiation and integration in a profound way. This theorem has two parts. The first part deals with finding antiderivatives, while the second part allows us to evaluate definite integrals using these antiderivatives.

The first part tells us that if a function is continuous over an interval, then it has an antiderivative over that interval. This antiderivative, in turn, can be used to compute the definite integral over that interval. This means that integration can be reversed by differentiation, and it allows us to find the net area under a curve.

The second part of the theorem is particularly handy for calculating definite integrals. It states that if you know an antiderivative of a function, you can evaluate the integral of that function from one point to another by taking the difference of the antiderivatives at those two points. In the context of this exercise, this is why we can substitute the antiderivative at the limits of integration to find the function \( F(x) \).

• The first part lays the groundwork, ensuring functions have antiderivatives if continuous.
• The second part directly provides a way to compute definite integrals using the antiderivatives.

Antiderivatives

Antiderivatives are essentially the reverse of derivatives. If you have a function, finding its antiderivative means you are looking for another function whose derivative is the original function.

In the exercise, we find the antiderivative of \( \frac{1}{\sqrt{t}} \). This is determined to be \( 2\sqrt{t} \). Thus, when differentiation is applied to \( 2\sqrt{t} \), it returns \( \frac{1}{\sqrt{t}} \).

This step is crucial because it allows us to replace the integral symbol with something tangible—a function expression—that we can manipulate. By finding the antiderivative, we transform the problem of finding an area or a cumulative value into a problem of simple arithmetic when evaluating at specific points.

• Antiderivatives help convert the integral into an evaluable form.
• They are found by reversing the differentiation process.
• Vital for solving definite integrals, they provide a concrete expression we can compute.

Variable Limit of Integration

Definite integrals with variable limits are a slightly more complex topic, but understanding them is key to mastering calculus applications.

In this context, a variable limit means that the upper (or lower) bound of integration isn’t a fixed number, but rather a function of another variable, like \( x^2 \) in the exercise. This means as \( x \) changes, the limit of integration changes, making the integral value dependent on \( x \).

The presence of variable limits requires us to carefully handle the integration process. Not only do we find an antiderivative, but we also need to substitute these limits into the antiderivative function. In our solution, \( x^2 \) was the variable upper limit, and after computing the antiderivative and substituting it, we simplified the result to express \( F(x) \) in terms of \( x \).

• Variable limits introduce functions as bounds in an integral.
• These require substitution into the antiderivative for evaluation.
• The result is a function of the variable, expressing the integral in terms of \( x \).