Problem 50 - Applied Calculus

Question

Find each integral. [Hint: Try some algebra.] $$ \int(x-1)^{2} \sqrt{x} d x $$

Step-by-Step Solution

Answer

The integral is \(\frac{2}{7}x^{3.5} - \frac{4}{5}x^{2.5} + \frac{2}{3}x^{1.5} + C\).

Step 1: Expand the Square

We start by expanding the squared term in the integrand. Using the formula \[(x - 1)^2 = x^2 - 2x + 1,\]we rewrite the integral as \[\int (x^2 - 2x + 1) \sqrt{x} \, dx.\]

Step 2: Distribute the Square Root

Next, distribute the \(\sqrt{x}\) to each term inside the integrand:\[\int (x^2\sqrt{x} - 2x\sqrt{x} + \sqrt{x}) \, dx.\]

Step 3: Simplify Each Term

Simplify each term by combining the powers of \(x\): - \(x^2\sqrt{x} = x^{2 + 0.5} = x^{2.5}\),- \(-2x\sqrt{x} = -2x^{1 + 0.5} = -2x^{1.5}\),- \(\sqrt{x} = x^{0.5}\).The integral becomes:\[\int (x^{2.5} - 2x^{1.5} + x^{0.5}) \, dx.\]

Step 4: Integrate Term by Term

Integrate each term individually:- \(\int x^{2.5} \, dx\): Increase the power by 1, giving \(x^{3.5}\), then divide by the new power: \[\int x^{2.5} \, dx = \frac{x^{3.5}}{3.5}.\]- \(\int -2x^{1.5} \, dx\): Similarly, this becomes: \[-2 \times \frac{x^{2.5}}{2.5} = -\frac{2x^{2.5}}{2.5}.\]- \(\int x^{0.5} \, dx\): This becomes: \[\frac{x^{1.5}}{1.5}.\]

Step 5: Combine the Results

Combine the integrals:\[\frac{x^{3.5}}{3.5} - \frac{2x^{2.5}}{2.5} + \frac{x^{1.5}}{1.5} + C,\]where \(C\) is the constant of integration. Simplify the fractions:\[\frac{1}{3.5}x^{3.5} - \frac{4}{5}x^{2.5} + \frac{2}{3}x^{1.5} + C.\]

Step 6: Final Simplified Form

Convert the coefficients to decimal or fraction form as needed:\[\frac{2}{7}x^{3.5} - \frac{4}{5}x^{2.5} + \frac{2}{3}x^{1.5} + C.\]

Key Concepts

Definite Integrals

Definite integrals are used to calculate the area under a curve described by a function over a specific interval. The general notation for a definite integral is \[ \int_{a}^{b} f(x) \ dx, \]where \(a\) and \(b\) are the bounds of integration, and \(f(x)\) is the function being integrated. When evaluating a definite integral, you find the antiderivative, evaluate it at the upper and lower bounds, and subtract the result derived at the lower bound from the upper. This gives a precise value, representing the net area under the curve.
Definite integrals are useful in a variety of applications, including determining distances, areas, and volumes in physical and engineering contexts. While our exercise focuses on indefinite integration, understanding definite integrals is crucial when you want to apply integration to real-world problems.

Indefinite Integrals

Indefinite integrals represent a family of functions whose derivatives give the integrand. The notation for an indefinite integral is \[ \int f(x) \, dx, \]where \(f(x)\) is the function you integrate. Unlike definite integrals, they don't have limits or bounds. Instead, they introduce a constant of integration, denoted as \(C\), since differentiating should account for any constant shifts in the function.
In our exercise, we integrated \( \int (x^{2.5} - 2x^{1.5} + x^{0.5}) \, dx, \)which resulted in specific antiderivatives,\( \frac{2}{7}x^{3.5} - \frac{4}{5}x^{2.5} + \frac{2}{3}x^{1.5} + C. \) These expressions are general solutions as they account for all possible functions that \( (x^{2.5} - 2x^{1.5} + x^{0.5}) \)could represent when differentiated.

Algebraic Manipulation

Algebraic manipulation is a crucial skill in integration, aiding in simplifying expressions into forms that are easier to integrate. In our example, we leveraged this skill to expand \((x-1)^2\)into \(x^2 - 2x + 1\), which is typically how you'd simplify polynomials.
Subsequently, we distributed the \(\sqrt{x}\)across the terms to get a simple sum of power terms, \(x^{2.5} - 2x^{1.5} + x^{0.5}, \)simplifying integration. Algebraic manipulation can also involve factoring, expanding, or rewriting expressions to help apply integration rules more efficiently. Being proficient with these techniques makes finding both indefinite and definite integrals easier when faced with complex expressions.

Power Rule for Integration

The Power Rule for Integration is a fundamental technique used to integrate functions of the form \(x^n\).To apply the power rule, you increase the exponent by one and then divide by the new exponent. The formula is \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \]where \(n eq -1\).
Our solution made use of this rule extensively:- For \(x^{2.5},\)we obtained\(\frac{x^{3.5}}{3.5} + C.\)- Similarly, applying the rule to\(-2x^{1.5}\)yielded\(-\frac{2x^{2.5}}{2.5}.\)- Finally, for \(x^{0.5},\)we ended up with\(\frac{x^{1.5}}{1.5}.\)The simplicity and utility of the power rule make it a staple in calculus, allowing you to handle polynomial functions easily through integration.