Problem 21 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int_{\sqrt{2}}^{1}\left(\frac{u^{7}}{2}-\frac{1}{u^{5}}\right) d u$$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{9}{8}\).

Step 1: Identify the Integral

We want to evaluate the integral \( \int_{\sqrt{2}}^{1}\left(\frac{u^{7}}{2}-\frac{1}{u^{5}}\right) d u \). This is a definite integral with the integrand \( \frac{u^{7}}{2} - \frac{1}{u^{5}} \) and limits of integration from \( \sqrt{2} \) to \( 1 \).

Step 2: Break Down the Integral

Separate the integral into two different integrals: \[ \int_{\sqrt{2}}^{1} \frac{u^{7}}{2} \, du - \int_{\sqrt{2}}^{1} \frac{1}{u^{5}} \, du. \]

Step 3: Integrate the First Term

Integrate \( \frac{u^{7}}{2} \): \[ \int \frac{u^{7}}{2} \, du = \frac{1}{2} \cdot \frac{u^8}{8} = \frac{u^8}{16}. \]

Step 4: Integrate the Second Term

Integrate \( -\frac{1}{u^5} \): \[ \int -u^{-5} \, du = \frac{u^{-4}}{4} = -\frac{1}{4u^4}. \]

Step 5: Evaluate the Antiderivatives at the Bounds

Now evaluate the antiderivatives from Steps 3 and 4 at the bounds \( \sqrt{2} \) and \( 1 \). Substitute \( 1 \) and subtract the expression evaluated at \( \sqrt{2} \).

Step 5.1: Evaluate the First Antiderivative

Substitute into \( \frac{u^8}{16} \): \[\left. \frac{u^8}{16} \right|_{\sqrt{2}}^1 = \frac{1^8}{16} - \frac{(\sqrt{2})^8}{16} = \frac{1}{16} - \frac{16}{16} = \frac{1}{16} - 1 = -\frac{15}{16}. \]

Step 5.2: Evaluate the Second Antiderivative

Substitute into \(-\frac{1}{4u^4}\): \[\left. -\frac{1}{4u^4} \right|_{\sqrt{2}}^1 = -\frac{1}{4(1)^4} + \frac{1}{4(\sqrt{2})^4} = -\frac{1}{4} + \frac{1}{16} = -\frac{4}{16} + \frac{1}{16} = -\frac{3}{16}. \]

Step 6: Combine the Results

Add the results from Steps 5.1 and 5.2: \(-\frac{15}{16} - \frac{3}{16} = -\frac{18}{16} = -\frac{9}{8}.\)

Key Concepts

Integration by Parts

Integration by parts is a technique used to integrate the product of two functions. While this method is powerful, it's not directly applicable to the integral given in this exercise. However, it's important to know that sometimes, when faced with a complex integral, integration by parts can be a valuable tool in the problem-solving toolbox. It is derived from the product rule of differentiation and is expressed by the formula:\[\int u \, dv = uv - \int v \, du\]In this context, it's useful for situations where the integral consists of products of algebraic, exponential, or trigonometric functions. The key to successfully applying integration by parts is selecting which part of the integrand to differentiate \(u\) and which to integrate \(dv\).
Although not used here, remember this technique as you tackle other integrals.

Antiderivatives

Finding antiderivatives, or indefinite integrals, is a fundamental part of evaluating definite integrals. In the given problem, we first need to find the antiderivatives of each part of the integrand separately.

• For the term \(\frac{u^7}{2}\), the antiderivative is \(\frac{u^8}{16}\).
• For \(-\frac{1}{u^5}\), rewrite it as \(-u^{-5}\) and find its antiderivative to be \(-\frac{1}{4u^4}\).

Antiderivatives can be viewed as the reverse process of differentiation. They return a function whose derivative is the original function being integrated. Therefore, determining an antiderivative is an essential step before applying limits of integration.

Limits of Integration

Once antiderivatives are found, the next step is to evaluate them at the bounds provided by the limits of integration. These limits define the interval over which the function is being integrated.
Here's the step-by-step process:

• Substitute the upper limit into the antiderivative.
• Substitute the lower limit into the antiderivative.
• Subtract the results of the lower limit from the upper limit evaluation.

In this exercise, the limits were \(\sqrt{2}\) and \(1\). After finding the antiderivatives, they were evaluated at these bounds. First, perform the substitution for each term separately and subtract, as demonstrated in the original solution, to combine the results. This final evaluation yields the value of the definite integral. Understanding how limits of integration work helps you apply the Fundamental Theorem of Calculus effectively.