Problem 47 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the definite integral. \(\int_{1}^{2} x \sqrt{x-1} d x\)

Step-by-Step Solution

Answer

The value of the definite integral is \(\frac{16}{15}\).

Step 1: Identify Substitution

Notice that the integrand has a composite function: \(x\sqrt{x-1}\). This suggests using substitution to simplify the integral. Let \(u = x - 1\). Then, \(du = dx\) and \(x = u + 1\).

Step 2: Change the Limits of Integration

Since \(u = x - 1\), the limits of integration need to be transformed. When \(x = 1\), \(u = 1 - 1 = 0\). When \(x = 2\), \(u = 2 - 1 = 1\). So, the new limits of integration are from 0 to 1.

Step 3: Substitute and Rewrite the Integral

Substitute \(u\) and \(du\) into the integral. The integral becomes: \[ \int_{0}^{1} (u + 1) \sqrt{u} ext{ } du \].

Step 4: Split the Integral

The integral \(\int_{0}^{1} (u + 1) \sqrt{u} ext{ } du \) can be split into two parts: \[ \int_{0}^{1} u \sqrt{u} ext{ } du + \int_{0}^{1} \sqrt{u} ext{ } du \].

Step 5: Simplify and Evaluate Each Integral

Rewrite \(u \sqrt{u}\) as \(u^{3/2}\) and \(\sqrt{u}\) as \(u^{1/2}\). Integrate each part separately:For \(\int_{0}^{1} u^{3/2} \underline{\phantom{xxx}} du\), use the power rule: \[ \frac{u^{5/2}}{5/2} \Bigg|_{0}^{1} = \frac{2}{5} \cdot 1^{5/2} - \frac{2}{5} \cdot 0 = \frac{2}{5} \].For \(\int_{0}^{1} u^{1/2} \underline{\phantom{xxx}} du\), use the power rule: \[ \frac{u^{3/2}}{3/2} \Bigg|_{0}^{1} = \frac{2}{3} \cdot 1^{3/2} - \frac{2}{3} \cdot 0 = \frac{2}{3} \].

Step 6: Combine the Results

Add the results of both integrals: \[ \frac{2}{5} + \frac{2}{3} \].Find a common denominator and combine:- \(\frac{2}{5} = \frac{6}{15}\)- \(\frac{2}{3} = \frac{10}{15}\)Thus, the combined result is:\[ \frac{6}{15} + \frac{10}{15} = \frac{16}{15} \].

Key Concepts

Integration Techniques

When faced with an integral, selecting the appropriate technique is crucial. Integration is the process of finding a function given its derivative. There are multiple techniques for evaluating integrals, and choosing the right one can make complex problems more manageable. Common methods include:

• Direct integration: Used if the integrand is straightforward.
• Substitution: Used for integrals with a composite function, useful in simplifying the problem.
• Partial fractions: Good for rational functions.
• Integration by parts: Useful when the integrand is a product of two functions.

In the exercise, the substitution method is particularly helpful because the integrand includes a composite function, making it complex to handle with direct approaches.

Substitution Method

The substitution method is often likened to reverse differentiation. It's handy when dealing with composite functions.To use this method, follow these steps:

• Choose a substitution: Pick a variable (usually "u") to simplify the integrand. For instance, if the integrand contains a form like \(x^2 + 1\), let \(u = x^2 + 1\).
• Rewrite the differential: After choosing \(u\), find the derivative \(du\) in terms of \(dx\).
• Change the limits (if it's a definite integral): If there are limits, adjust them according to the substitution.
• Substitute and integrate: Replace variables in the integrand with the new ones involving \(u\) and \(du\). Then, perform the integration.

In the original problem, \(u = x - 1\) simplifies the term \(\sqrt{x-1}\), which makes the integration process much smoother.

Limits of Integration

The limits of integration define the interval over which you integrate a function. They are crucial in the evaluation of definite integrals as they specify the start and end points of integration. For definite integral evaluation:

• Transform the limits with substitution: When substitution is used, the original limits need to be converted to match the substitution. Calculate the new limits by plugging the old ones into your substitution equation.
• Evaluate the antiderivative at these limits: Once integrated, the result is evaluated at the transformed limits to find the definite integral's value.

In the example, the initial problem had limits of 1 and 2. Substitution adjusted these to 0 and 1, creating a new framework for evaluating the integral.

Power Rule in Integration

The power rule is perhaps the most frequently used tool in integration. It helps when integrating polynomial functions.The rule states:

• The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), whenever \(n eq -1\).
• This rule is beautifully simple and works well on terms with power functions.

In the step-by-step breakdown, the integrals \(\int u^{3/2} \, du\) and \(\int u^{1/2} \, du\) both utilize the power rule. They evaluate directly to expressions involving \(u^{5/2}\) and \(u^{3/2}\), respectively. Applying this rule results in easy calculation of definite integrals, once the limits of integration have been substituted.