Problem 12 - Essential Calculus

Question

Find the indefinite integral and check the result by differentiation. $$ \int\left(4 x^{3}+6 x^{2}-1\right) d x $$

Step-by-Step Solution

Answer

The indefinite integral of \( u^{2} \sqrt{u^{3}+5} \, du \) is \( (u^3 + 5)^{3/2}/2 + C \).

Step 1 - Setup

The expression is \( \int u^{2} \sqrt{u^{3}+5} \, du \). Start by letting \( v = u^3 + 5 \). This looks promising because the derivative of \( v \) with respect to \( u \) would give \( 3u^2 \), which is part of the integrand.

Step 2 - Differentiate and Substitution

Differentiate \( v \) with respect to \( u \) to get \( dv = 3u^2 du \). From this, we can express \( du \) in terms of \( dv \) which is \( du = dv / (3u^2) \). Substitute \( v = u^3 + 5 \) and \( du = dv / (3u^2) \) into the integral to get \( (1/3) \int \sqrt{v} \, dv \).

Step 3 - Solve the Integral

Now, compute the integral. The antiderivative of \( \sqrt{v} \) is \( (2/3)v^{3/2} \), so \( (1/3) \int \sqrt{v} \, dv = (1/3) * (2/3) * v^{3/2} = v^{3/2}/2 \).

Step 4 - Substitute \( u \) Back

Substitute \( u^3 + 5 \) back for \( v \) to get \( (u^3 + 5)^{3/2}/2 + C \), where \( C \) is the constant of integration.

Step 5 - Check by Differentiation

To check the result, differentiate \( (u^3 + 5)^{3/2}/2 \). Using the chain rule, we get \( u^2 \sqrt{u^3 + 5} \), which confirms the original expression.

Key Concepts

Integration by Substitution

Integration by substitution is a fundamental technique used to simplify finding integrals, just like we sometimes substitute variables in algebra to solve equations. It's especially useful when dealing with composite functions, which feature a function within a function.
The core idea is to choose a substitution that makes the integral easier to solve. In many cases, this involves swapping the variable of integration for a new one that simplifies the expression.

• Begin by identifying the part of the integrand that can be a derivative after substitution. In our exercise, selecting \( v = u^3 + 5 \) simplifies our work because the derivative \( dv = 3u^2 du \) aligns with the original integrand.
• Next, express \( du \) in terms of \( dv \) so you can perform the substitution effectively. Here, \( du = dv / (3u^2) \).
• Replace the variables in the integral with these new expressions, moving from more complex to simpler functions.

Antiderivative

The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. When we integrate a function, we are essentially finding a new function whose derivative is the one we started with. This is a crucial concept in calculus for solving a wide variety of problems.
In our exercise, after performing the substitution, we simplified the integral to \( \frac{1}{3} \int \sqrt{v} \, dv \). Finding the antiderivative of \( \sqrt{v} \) involves recognizing its derivative. You apply the power rule in reverse:

• Express \( \sqrt{v} \) as \( v^{1/2} \).
• Add one to the exponent: \( 1/2 + 1 = 3/2 \).
• Divide by the new exponent to adjust for the antiderivative: \( 2/3 \cdot v^{3/2} \).

Combining these steps with the substitution gives us the antiderivative in terms of \( v \). This is then converted back to \( u \) to provide the full solution.

Chain Rule

The chain rule is a key concept in calculus used to differentiate composite functions. It says that if you have a function within another function, you differentiate the outer function while respecting the derivative of the inner function. This rule helps track the rate of change in layered phenomena.
To check our integral result, we re-apply the chain rule to differentiate our solution \( (u^3 + 5)^{3/2}/2 \). Here's how:

• The outer function is \( (v)^{3/2} \) where \( v = u^3 + 5 \). Differentiate with respect to \( v \) first.
• Now differentiate the inner part \( u^3 + 5 \) to get \( 3u^2 \).
• This yields \( 3/2 \cdot (u^3 + 5)^{1/2} \times 3u^2 \), capturing the original function \( u^2 \sqrt{u^3 + 5} \).

Differentiation

Differentiation is one of the principal operations in calculus, employed to ascertain how a function changes at any given point. Essentially, it determines the rate at which a function's values are changing over time or in relation to another variable.
When we differentiate, we calculate the slope of the tangent line at any point on the graph of a function, breaking down complex behaviors into manageable data points.

• In our context, differentiation was our tool for verifying the correctness of our integration work, ensuring the derivative of our solution returned to the original function.
• This step acts as both a theoretical and practical confirmation of our integration process.

By confirming the derivative leads us back to the starting function, we cemented our understanding and solution accuracy for the integral problem.