Problem 66 - Calculus Early Transcendentals

Question

Simplify the following expressions. $$\frac{d}{d x} \int_{x}^{0} \frac{d p}{p^{2}+1}$$

Step-by-Step Solution

Answer

Answer: The simplified expression is $$-\frac{1}{x^2 + 1}$$.

Step 1: Identify the integrand

The integral here is $$\int_{x}^{0} \frac{d p}{p^{2}+1}$$. The integrand inside the integral is: $$\frac{1}{p^2 + 1}$$.

Step 2: Apply the Fundamental Theorem of Calculus (FTC)

Recall the FTC: If F'(x) = f(x) then $$\frac{d}{d x} \int_{a(x)}^{b(x)} f(p) dp = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$$. In our case, the integral is from x to 0, so $$a(x) = x$$ and $$b(x) = 0$$. Therefore, $$a'(x) = 1$$ and $$b'(x) = 0$$. Now, we can apply the FTC:$$\frac{d}{d x} \int_{x}^{0} \frac{d p}{p^{2}+1} = \frac{1}{0^2 + 1} \cdot 0 - \frac{1}{x^2 + 1} \cdot 1$$

Step 3: Simplify the result

We got: $$\frac{1}{0^2 + 1} \cdot 0 - \frac{1}{x^2 + 1} \cdot 1$$Now, let's simplify:$$0 - \frac{1}{x^2 + 1} = -\frac{1}{x^2 + 1}$$ Therefore, the simplified expression is:$$-\frac{1}{x^2 + 1}$$.

Key Concepts

Integrand

When we talk about an integrand, we are referring to the function that is being integrated in an integral. In the context of our exercise, the integrand is the mathematical expression \( \frac{1}{p^2 + 1} \). It is crucial to identify the integrand correctly since it is the foundation for applying integration techniques and understanding the behavior of the integral.

Understanding the role of the integrand is also helpful in simplifying expressions within the integral. For instance, if the integrand can be algebraically manipulated to a known form, it might make the integration process or its subsequent differentiation much more straightforward.

Differentiation Under the Integral Sign

The process called differentiation under the integral sign is a technique where we take the derivative of an integral with respect to an external variable. This concept is closely tied to the Fundamental Theorem of Calculus, which provides a connection between differentiation and integration.

In our exercise, the differentiation was performed under the integral sign with respect to variable \( x \). Here's the important thing: when the limits of the integral depend on the variable we are differentiating with respect to, we must take into account the rate at which these limits change, which refers to their derivatives.

Simplifying Expressions

The art of simplifying expressions is vital to effectively communicate the essence of mathematical problems. In our exercise, after the differentiation under the integral sign, the expression obtained is

\[0 - \frac{1}{x^2 + 1} = -\frac{1}{x^2 + 1}\]

. This simplification involves recognizing that multiplying by zero yields zero and eliminates terms that do not contribute to the final result.

Understanding how to simplify expressions simplifies problem-solving, reduces the potential for errors, and paves the way to clearer insights into the problems at hand. For example, noticing that one part of the expression becomes simply zero due to multiplication allows for faster recognition of the solution.

Derivative of an Integral

The derivative of an integral concept comes from the Fundamental Theorem of Calculus, which helps us to directly compute the derivative without the need for performing actual integration. In the given problem, we are looking at the derivative of an integral with variable limits. The theorem tells us how to relate this derivative to the function inside the integral — the integrand.

Here, the derivative of the integral from \( x \) to \( 0 \) of \( \frac{1}{p^2 + 1} \) with respect to \( x \) reduces to the integrand evaluated at \( x \), with a change of sign. This is because the upper limit is a constant (and hence its derivative is zero), while the lower limit is a variable function whose derivative is \(1\). This connects the seemingly separate operations of differentiation and integration in a concrete and useful way.