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Chapter 5: Problem 32

Evaluate the integrals by any method. $$\int_{1}^{2} \sqrt{5 x-1} d x$$

Short Answer

Expert verified
The integral evaluates to \( \frac{38}{15} \).

Step by step solution

01

Choose a Substitution

To evaluate the integral \( \int_{1}^{2} \sqrt{5x-1} \, dx \), a substitution method is appropriate. Let's choose \( u = 5x - 1 \). Then the derivative is \( du = 5 \, dx \), which implies \( dx = \frac{1}{5} \, du \).
02

Change the Limits of Integration

Since we have a substitution, we need to change the limits according to the new variable \( u \). When \( x = 1 \), \( u = 5(1) - 1 = 4 \). When \( x = 2 \), \( u = 5(2) - 1 = 9 \). Thus, the new limits of integration are from 4 to 9.
03

Substitute and Transform the Integral

Substitute \( u \) and \( dx = \frac{1}{5} \, du \) into the integral to get: \[ \int_{4}^{9} \sqrt{u} \cdot \frac{1}{5} \, du = \frac{1}{5} \int_{4}^{9} u^{1/2} \, du. \]
04

Integrate the Transformed Integral

Integrate \( \int u^{1/2} \, du \) using the power rule, which gives: \[ \frac{1}{5} \left[ \frac{u^{3/2}}{3/2} \right]_{4}^{9}. \] This simplifies to \( \frac{1}{5} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{4}^{9} \).
05

Evaluate the Definite Integral

Compute the expression at the transformed limits: \( \frac{2}{15} \left[ (9)^{3/2} - (4)^{3/2} \right] \). Calculate \( 9^{3/2} = 27 \) and \( 4^{3/2} = 8 \), resulting in \( \frac{2}{15} (27 - 8) = \frac{2}{15} \times 19 = \frac{38}{15} \).
06

Final Answer

The evaluated integral is \( \frac{38}{15} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful tool for evaluating integrals, especially when dealing with more complicated functions. The idea is to simplify the integral by substituting a part of it with a new variable. In our exercise, we choose the substitution \( u = 5x - 1 \). This choice is strategic because it transforms the integral into a simpler form. By solving for \( dx \) with the substitution equation, \( dx = \frac{1}{5} \, du \), we adjust the integral accordingly.
  • Choosing \( u \) correctly can make or break the ease of solving an integral, as it simplifies the expression.
  • Always take the derivative of \( u \) to express \( dx \) in terms of \( du \).
The substitution not only simplifies the integrand but also makes it easier to apply the power rule later. Keep in mind that any time you choose a substitution, you must also change the integration limits to match the new variable!
Power Rule in Integration
The power rule in integration is one of the fundamental techniques for finding the integral of a function. This rule states that for any function \( u^n \), the integral is given by:\[\int u^n \, du = \frac{u^{n+1}}{n+1} + C\]where \( C \) is the constant of integration. In our exercise, the integrand after substitution is \( u^{1/2} \).
  • Applying the power rule, \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} \), which simplifies nicely.
  • Remember, the power rule only works if \( n eq -1 \).
Evaluating this, and multiplying by the constant from substitution (\( \frac{1}{5} \)), yields \( \frac{1}{5} \cdot \frac{2}{3} [u^{3/2}]_{4}^{9} \). The power rule makes integration an easier and straightforward process when applied correctly.
Limits of Integration
When performing a substitution in a definite integral, changing the limits of integration is critical. These new limits must be in terms of the variable you've substituted to. In the given exercise, when \( x = 1 \), the corresponding \( u \) value is calculated as \( u = 5(1) - 1 = 4 \). Similarly, when \( x = 2 \), \( u = 5(2) - 1 = 9 \). Thus, the limits transform from \( x = 1 \) and \( x = 2 \) to \( u = 4 \) and \( u = 9 \).
  • Alter the limits using the substitution formula to get them in terms of \( u \).
  • This step is necessary for completing the integral within the specified boundaries.
Incorrect limits can lead to wrong answers, but by carefully computing the new limits, the evaluation becomes accurate, as seen when we substitute and solve the transformed integral to get \( \frac{38}{15} \). Understanding this shift in limits ensures no part of the original function's behavior is lost in the integration process.

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Most popular questions from this chapter

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$V=V_{p} \sin (2 \pi f t)$$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts (V), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). (1 \(\mathrm{Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the rms or root -mean-square value of \(V .\) By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}}$$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f,\) and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral. \(]\) (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{V}\) at a frequency of \(60 \mathrm{Hz}\). What is the peak voltage at such an outlet? (Check your book to see figure)

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right)$$

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Evaluate each integral by first modifying the form of the integrand and then making an appropriate substitution, if needed. $$\int \frac{t+1}{t} d t$$
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