Problem 49 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indefinite integral. $$\int v^{2}(1-v)^{6} d v$$

Step-by-Step Solution

Answer

The indefinite integral of \(v^2 (1-v)^6\) with respect to \(v\) is \(\int v^{2}(1-v)^{6} d v = -(\frac{(1-v)^9}{9} - 2\frac{(1-v)^8}{8} + \frac{(1-v)^7}{7}) + C\), where C is the constant of integration.

Step 1: Apply substitution method

Let \(u = 1-v\), which means \(v = 1-u\). Now we need to find \(d u\) with respect to \(v\), and replace it in the integral.

Step 2: Find the derivative of u with respect to v

Differentiate \(u\) with respect to \(v\): \( \frac{d u}{d v} = -1 \) Now, multiply both sides by \(d v\), we have: \( d u = -d v \)

Step 3: Substitute u and dv in the integral

Now we can replace \(v\) and \(d v\) in the integral: \(\int v^{2}(1-v)^{6} d v = \int (1-u)^{2}(u)^{6} (-d u) \)

Step 4: Simplify the integral

We have: \(\int (1-u)^{2}(u)^{6} (-d u) = -\int (1-2u+u^{2})(u^{6}) d u \)

Step 5: Expand the integrand

Now we need to expand the integrand: \(-\int(1-2u+u^{2})(u^{6}) d u = -\int(u^{8}-2u^{7}+u^{6}) d u\)

Step 6: Integrate term by term

Now we can integrate each term: \(-\int(u^{8}-2u^{7}+u^{6}) d u = -(\frac{u^9}{9} - 2\frac{u^8}{8} + \frac{u^7}{7}) + C\) Where \(C\) is the constant of integration.

Step 7: Substitute back for v

Now we substitute back for \(v\): \(-(\frac{u^9}{9} - 2\frac{u^8}{8} + \frac{u^7}{7}) + C = -(\frac{(1-v)^9}{9} - 2\frac{(1-v)^8}{8} + \frac{(1-v)^7}{7}) + C\) So the indefinite integral is \(\int v^{2}(1-v)^{6} d v = -(\frac{(1-v)^9}{9} - 2\frac{(1-v)^8}{8} + \frac{(1-v)^7}{7}) + C\).

Key Concepts

Substitution Method

Mastering the substitution method is like having a secret decoder ring for solving integrals. It's an elegant technique often used when an integral contains a composite function. Imagine you're working on a complex jigsaw puzzle, and substitution is your strategy for piecing together a tricky section by simplifying it into more manageable parts.

Think of the substitution method as a transformation; you're temporarily changing the variables to make the integral look simpler. The goal is to choose a new variable, typically represented by 'u,' to replace a part of the original integral. This new variable 'u' simplifies the expression and, importantly, we need to find the differential 'du' which represents the rate at which 'u' changes with respect to the original variable.

The real magic happens when you swap out sections of the original integral with these new 'u' terms. Suddenly, your intimidating integral is now an easier-to-handle expression, which you can then integrate with more confidence. Always remember, though, the final step in the substitution method is to convert back to the original variable, essentially reversing the initial transformation, to ensure the solution is in terms of the original problem's variable.

Integration by Parts

Integration by parts is essentially the reverse gear of the product rule of differentiation, enabling us to integrate products of functions. It's the mathematical equivalent of teamwork; each part works with the other to achieve the goal of integration. But it's not just about teamwork; knowing when and how to use this method is crucial.

Integration by parts can be summarized by the formula \[ \int u dv = uv - \int v du \.\] The strategy involves choosing which part of the function to differentiate (making it 'du') and which part to integrate (making it 'dv'). Usually, the choice is made to simplify the integral, making one part easier to integrate while the other becomes a simpler derivative.

The beauty of integration by parts is in breaking down a complex integral into simpler parts. It's a powerful tool—but choose your 'u' and 'dv' wisely, as the wrong choice can make the problem more complex rather than less. Once you've integrated by parts, sometimes you might need to repeat the process several times, or even combine it with other techniques—such as the substitution method—to fully solve the integral.

Constant of Integration

The constant of integration can be considered the mysterious chameleon of calculus. It's the symbol 'C' that appears in the solution of indefinite integrals, and it represents an entire family of possible functions that differ only by a constant. When you integrate, you're finding all possible antiderivatives, and this constant reminds us that integration is not a one-way street; there are many possible paths that lead back to the original function if we were to differentiate it again.

Why is there a constant of integration? When you take a derivative, information about any added or subtracted constant vanishes, like footprints washed away by the tide. When we reverse the process through integration, we account for that lost information by including 'C', acknowledging all the possible starting points we could have had before differentiation. It keeps the integrity of calculus intact by recognizing the infinite possibilities that exist within a simple integral. Always include the constant 'C' in your indefinite integrals to express the full breadth of solutions and to showcase your deep understanding of integral calculus.