Problem 42 - Calculus

Question

Evaluate by any method. $$ \frac{d}{d x} \int_{x}^{x^{2}} \frac{d t}{t} $$

Step-by-Step Solution

Answer

The solution to the integral's derivative is \( \frac{1}{x} \).

Step 1: Understand the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus relates differentiation and integration, and it states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x) \). However, in this problem, both limits of integration depend on \(x\), requiring a different technique.

Step 2: Identify the Leibniz Rule for Differentiation of Integrals

The Leibniz rule helps differentiate integrals with variable limits. If \( F(x) = \int_{u(x)}^{v(x)} f(t) \, dt \), then \( \frac{dF}{dx} = f(v(x))v'(x) - f(u(x))u'(x) \). Here, \( u(x) = x \) and \( v(x) = x^2 \).

Step 3: Set Up the Application of Leibniz Rule

Write the integrand \( f(t) = \frac{1}{t} \). For \( u(x) = x \), we have \( u'(x) = 1 \), and for \( v(x) = x^2 \), we get \( v'(x) = 2x \). Substitute these into the Leibniz rule equation.

Step 4: Apply the Leibniz Rule

Substituting into the Leibniz rule, we have: \( \frac{1}{v(x)}v'(x) - \frac{1}{u(x)}u'(x) = \frac{1}{x^2} \cdot 2x - \frac{1}{x} \cdot 1 = \frac{2x}{x^2} - \frac{1}{x} \).

Step 5: Simplify the Expression

Simplify the expression from the previous step: \( \frac{2x}{x^2} = \frac{2}{x} \) and \( \frac{1}{x} = \frac{1}{x} \). Thus, \( \frac{2}{x} - \frac{1}{x} = \frac{1}{x} \).

Step 6: Finalize the Solution

The simplified result represents the derivative of the integral with respect to \( x \). So, the final answer is \( \frac{1}{x} \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a vital bridge connecting the concepts of differentiation and integration. It is divided into two main parts:

• The first part tells us how to evaluate definite integrals by relating it to the antiderivative, while the second part essentially says that the derivative of an integral will give us back the original function.

In its simplest form, it is expressed as: if a function \(f(t)\) is continuous on the interval \([a, b]\), then the function \(F(x) = \int_{a}^{x} f(t)\,dt\) is continuous on \([a, b]\) and differentiable on \((a, b)\), and \( F'(x) = f(x) \).
In more practical terms, this theorem showcases that calculating the definite integral of a function over an interval can be accomplished using one of its antiderivatives. However, the exercise deals with variable limits, necessitating a more advanced method than the basic application of the Fundamental Theorem.

Differentiation under the Integral Sign

Differentiation under the integral sign, often associated with the Leibniz rule, is a powerful technique for evaluating integrals where the boundaries of integration are functions of a variable. This approach is particularly useful when both the upper and lower limits depend on the variable in question, as seen in this exercise.

• The basic idea is that if you have an integral whose limits are functions of \(x\), the derivative of that integral with respect to \(x\) involves evaluating the integrand at the boundary limits, followed by multiplying it by the derivative of the respective limits.

This rule is rarely used in simpler calculus problems but shines in more complex scenarios. For instance, differentiating an integral \(\int_{u(x)}^{v(x)} f(t) \, dt\) with variable limits allows you to discover solutions otherwise elusive without directly solving the integral with respect to time.

Variable Limits of Integration

In calculus, variable limits of integration are crucial for dealing with problems where the boundaries of an integral are dependent on another variable. In this exercise, both the upper and lower limits of the integral are functions of \(x\).
When evaluating such integrals, normal procedures differ because the limits are not constants. The Leibniz rule provides a systematic method to manage this complexity. It recognizes that the derivative with respect to \(x\) of an integral with limits \(x\) and \(x^2\) will factor in both boundary derivatives. So, in our exercise, for limits \(u(x) = x\) and \(v(x) = x^2\), their derivatives, \(u'(x) = 1\) and \(v'(x) = 2x\), respectively, were used with the integrand \(f(t)= \frac{1}{t}\) to simplify to the result \(\frac{1}{x}\).

• This method helps navigate complex problems that involve functions as integration limits instead of constants, maintaining analytical simplicity despite the involvement of non-static boundaries.

Understanding how variable limits influence integral evaluation is a foundational skill in advanced calculus, enabling precise manipulation and differentiation of functions defined across dynamic intervals.