Problem 28 - Calculus and its applications

Question

Evaluate. (Be sure to check by differentiating!) $$ \int\left(x^{3}-x^{2}-x\right)^{9}\left(3 x^{2}-2 x-1\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{(x^3 - x^2 - x)^{10}}{10} + C \).

Step 1: Identify a Suitable Substitution

To solve the integral, we'll use substitution. Let's set the inner function that is raised to the power of 9 as the substitution: let \( u = x^3 - x^2 - x \). Then, compute the derivative: \( \frac{du}{dx} = 3x^2 - 2x - 1 \).

Step 2: Perform the Substitution

Replace \( x^3 - x^2 - x \) with \( u \) and \( 3x^2 - 2x - 1 \) with \( \frac{du}{dx} \) in the integral. This transforms the integral to \( \int u^9 \, du \).

Step 3: Integrate with Respect to u

Now, integrate \( u^9 \) with respect to \( u \). The integral of \( u^9 \) is \( \frac{u^{10}}{10} + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back in Terms of x

Substitute \( u = x^3 - x^2 - x \) back into the integral result. This gives the antiderivative: \( \frac{(x^3 - x^2 - x)^{10}}{10} + C \).

Step 5: Check by Differentiating the Result

Differentiate the found antiderivative \( \frac{(x^3 - x^2 - x)^{10}}{10} + C \) to verify it matches the original integrand \( (x^3 - x^2 - x)^9 (3x^2 - 2x - 1) \). Using the chain rule, the derivative of \( \frac{(x^3 - x^2 - x)^{10}}{10} \) is \( (x^3 - x^2 - x)^9 (3x^2 - 2x - 1) \), confirming the integration is correct.

Key Concepts

Substitution Method

The substitution method is a powerful technique in integral calculus. It helps simplify complex integrals by transforming them into simpler forms, making them easier to solve. Think of it like changing variables in an equation to make it more manageable.
For example, if we encounter the integral \( \int f(g(x))g'(x) \,dx \), we can use substitution. We set \( u = g(x) \), which transforms our integral, making it respect to \( u \) instead of \( x \). The derivative \( \frac{du}{dx} = g'(x) \) helps us express \( dx \) in terms of \( du \).

• Identify a part of the integrand (function inside) to substitute for \( u \).
• Compute \( du \) by differentiating \( u \).
• Transform the integral into one in terms of \( u \).
• Integrate the simpler expression in terms of \( u \).
• Finally, substitute back to get the result in original variables.

In our example, we let \( u = x^3 - x^2 - x \), simplifying our complex integral to \( \int u^9 \, du \).

Chain Rule

The chain rule is a fundamental concept in calculus, pivotal for finding derivatives of composite functions. It's like peeling away layers to reach the center, allowing us to differentiate complex compositions effectively.
When you have a function \( y = f(g(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). In simple terms, this means you differentiate the outer function and multiply it by the derivative of the inner function.
During integration, especially when checking our solution by differentiation (as we often do with substitution), the chain rule ensures the form and structure align with the original function. When we differentiate our antiderivative \( \frac{(x^3 - x^2 - x)^{10}}{10} \), we apply the chain rule:

• The power \( 10 \) comes down, reducing the power to \( 9 \).
• Multiply by the derivative of the inside, \( 3x^2 - 2x - 1 \).

This confirms our solution matches the original integral expression.

Antiderivative

Antiderivatives, sometimes referred to as indefinite integrals, are the 'reversal' of taking a derivative. It's akin to understanding the original function from its rate of change.
Finding an antiderivative is like guessing the function that gives the known derivative. Knowing basic rules and transformations are key to successful guesses.
For the power rule, which is a frequent method in finding antiderivatives, if \( F'(x) = x^n \), then \( F(x) = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
In our problem, after substitution, we seek the antiderivative of \( u^9 \). By applying the power rule, we find the antiderivative to be \( \frac{u^{10}}{10} + C \). When we substitute back, it yields the final result in terms of \( x \).
Checking by differentiating this expression proved fruitful, suggesting success in our integration method.