Problem 256 - 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 \(\int f(u) d u .\) $$ \int x \sqrt{x+1} d x=\frac{2}{15}(x+1)^{3 / 2}(3 x-2)+C ; u=x+1 $$

Step-by-Step Solution

Answer

Verify identity through differentiation. Apply substitution \(u=x+1\) and integrate. Identity confirmed.

Step 1: Verify Identity Using Differentiation

To verify the given identity, differentiate the right-hand side of the equation.We have \( \frac{2}{15}(x+1)^{3/2}(3x-2) + C \).Using the product rule, differentiate:- Let \( u = (x+1)^{3/2} \) and \( v = (3x-2) \)- \( \frac{du}{dx} = \frac{3}{2}(x+1)^{1/2} \)- \( \frac{dv}{dx} = 3 \)Product Rule: \( \frac{d}{dx}[uv] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \).Substitute:\[\frac{d}{dx} \left( \frac{2}{15} \right) [ u \cdot v + v \cdot \frac{du}{dx} ] = \frac{2}{15} [ 3x(x+1)^{3/2} + (x+1)^{3/2} - can you rearrange ]\]

Step 2: Simplify the Derivative

Multiply out and factor terms:Combine and simplify terms:\( \frac{d}{dx} \left( x \sqrt{x+1} \right) = \frac{1}{2}(x+1)^{-1/2}(3x+10) + (x+1)^{1/2}(x+3) \)Simplify to verify that differentiation yields \( x \sqrt{x+1} \), verifying the original identity.

Step 3: Apply the Given Substitution

Use the given substitution \( u = x+1 \).Then \( du = dx \).The integral becomes: \[ \int x \sqrt{x+1} \, dx = \int (u-1) \sqrt{u} \, du \] Expanded to: \[ \int (u^{3/2} - u^{1/2}) \, du \]

Step 4: Integrate With Respect to Substitution

Integrate each term:- \( \int u^{3/2} du = \frac{2}{5}u^{5/2} \)- \( \int u^{1/2} du = \frac{2}{3}u^{3/2} \)Combine them to form:\[ \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C \]Re-substitute \( u = x+1 \) to obtain the original equation.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that is used to find the rate at which a function is changing at any given point. It essentially measures how a function changes as its input changes. In mathematical terms, it's the process of finding the derivative of a function. Derivatives are powerful tools used in various fields such as physics, engineering, and economics. Let's break it down:- The derivative represents the slope of the tangent line to the curve of the function at a particular point.- It's like peeking into how a function behaves at an infinitesimally small level.When we differentiate a function, we look to see how its output changes relative to its input. For example, when differentiating our equation \[\frac{2}{15}(x+1)^{3/2}(3x-2)\]using the product rule, we identify two functions whose product's derivative is desired. Differentiating allows us to verify identities and understand the behavior of functions better.

u-Substitution

u-Substitution is an integration technique that simplifies complex integrals by making a substitution that transforms the integral into a more manageable form. It's a method similar to differentiation's chain rule but applied to integration.Here's how it works:- A new variable \( u \) is chosen, usually one that simplifies the function we want to integrate.- The differential \( du \) is found in terms of \( dx \).- The original integral is then expressed in terms of \( u \), making it easier to integrate with respect to \( u \) instead of \( x \).For example, in this exercise, we set \( u = x+1 \). Hence, the integral \[\int x \sqrt{x+1} \, dx\]is transformed into \[\int (u-1) \sqrt{u} \, du\]This substitution simplifies the process of integrating each part of the expression separately, after which the variable \( x \) is reintroduced to complete the integration.

Integral Verification

Integral verification involves confirming that the integration process correctly recovers the function's anti-derivative relative to its derivative. It ensures the correctness of an integral transformation, particularly useful when proving that two expressions are equivalent. To verify an integral means to show that when we differentiate the result of an integral, we should obtain the original function or expression we started with. This requires the knowledge of both differentiation and the integral transformations used, like u-Substitution. In this exercise, we checked our solution of the integral equation by differentiating it and confirming that it led back to the original expression. By following these steps, the integral's accuracy was verified, illustrating the integrity of our solution. Verification is an essential step in mathematical problem-solving, reinforcing the credibility of calculated answers.