Problem 258 - Calculus Volume 1

Question

In the following exercises, verify each identity using differentiation. Then, using the indicated \(u\) -substitution, identify \(f\) such that the integral takes the form \(\int f(u) d u .\) $$ \int x \sqrt{4 x^{2}+9} d x=\frac{1}{12}\left(4 x^{2}+9\right)^{3 / 2}+C ; u=4 x^{2}+9 $$

Step-by-Step Solution

Answer

The identity is verified, and with \( u = 4x^2 + 9 \), we have \( f(u) = \frac{1}{8}\sqrt{u} \).

Step 1: Differentiate the Right Side

We begin by differentiating the right-hand side of the given identity. The expression is \( \frac{1}{12}(4x^2 + 9)^{3/2} + C \). Differentiate it with respect to \(x\): \[ \frac{d}{dx} \left( \frac{1}{12}(4x^2 + 9)^{3/2} \right) = \frac{1}{12} \cdot \frac{3}{2} \cdot (4x^2 + 9)^{1/2} \cdot 8x. \] Simplifying this, we get \( x \cdot \sqrt{4x^2 + 9} \). This matches the integrand, verifying the identity.

Step 2: Identify the Function for Integration

Given the substitution \( u = 4x^2 + 9 \), we need to express the integral in terms of \( u \). Start by differentiating \( u \) to find \( du \): \( du = 8x \, dx \). Thus, \( dx = \frac{du}{8x} \). Now, rewrite the integral: \[ \int x \sqrt{4x^2 + 9} \, dx = \int x \sqrt{u} \frac{du}{8x} \] which simplifies to \[ \frac{1}{8} \int \sqrt{u} \, du. \] Let \( f(u) = \frac{1}{8} \sqrt{u} \), so the integral takes the form \( \int f(u) \, du \).

Key Concepts

Differentiation

Differentiation is a fundamental aspect of calculus that involves finding the derivative of a function. It is crucial because it tells us how a function changes as its input changes. Here, we use differentiation to verify an identity in an integral problem.

Given a composite function like \( \frac{1}{12}(4x^2 + 9)^{3/2} + C \), we apply the chain rule to differentiate it. The chain rule states that if you have a function of a function, then its derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function itself.

• Start by differentiating the outer function: \( \frac{d}{dz} z^{3/2} = \frac{3}{2}z^{1/2} \).
• Next, multiply by the derivative of the inner function: \( \frac{d}{dx}(4x^2 + 9) = 8x \).

Putting these together, the derivative becomes \( \frac{1}{12} \cdot \frac{3}{2} \cdot (4x^2 + 9)^{1/2} \cdot 8x \), which simplifies to \( x \cdot \sqrt{4x^2 + 9} \), therefore confirming that the differentiation is correct. This matches the original integrand and establishes our confidence that the original problem's identity holds.

Integration

Integration is another key concept in calculus, which is the reverse process of differentiation. It involves finding a function given its derivative, often used to find areas under curves.

In this exercise, we are asked to verify an integral equation, meaning we must ensure that its integration leads us back to the original function given after the integral sign. The way to achieve this is often to apply substitution methods that can simplify complex functions to more manageable forms.

• In our case, the integration began with the function \( x \sqrt{4x^2 + 9} \), which may appear daunting at first.
• However, by considering a suitable substitution, we transition from a complex integrand to a form that is easier to integrate.

The integration ultimately shows that the original function can indeed be represented by the simple process of differentiating the right-hand expression confirmed by differentiation earlier, thereby verifying its identity.

u-substitution

The method of u-substitution is a vital tool for solving integrals, especially when dealing with composite functions. It's like undoing the chain rule used in differentiation, allowing you to simplify the function into basic components.

Here's how it works: you replace a part of the integrand with a new variable, \( u \), to make integration easier. In our exercise, we choose \( u = 4x^2 + 9 \). This substitution changes the integrand \( x \sqrt{4x^2 + 9} \) into something simpler.

• Differentiate \( u \) to find \( du \), which gives \( du = 8x \, dx \). Therefore, \( dx = \frac{du}{8x} \).
• Insert these into the integral: \( \int x \sqrt{u} \frac{du}{8x} \), which simplifies to \( \frac{1}{8} \int \sqrt{u} \, du \).

Now, the problem has become a basic integral with a straightforward solution \( f(u) = \frac{1}{8} \sqrt{u} \). Hence, u-substitution neatly converts a complicated integral into one that is much simpler to solve, illustrating its efficiency and utility in calculus.