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\(21-44\) Evaluate the integral. $$\int_{1}^{9} \frac{3 x-2}{\sqrt{x}} d x$$

Short Answer

Expert verified
The integral evaluates to 26.

Step by step solution

01

Rewrite the Integrand

Let's start by rewriting the integrand to make it easier to integrate. Given the integral \( \int_{1}^{9} \frac{3x-2}{\sqrt{x}} \, dx \), rewrite \( \frac{3x-2}{\sqrt{x}} \) as \( 3x \cdot x^{-1/2} - 2x^{-1/2} \) which simplifies to \( 3x^{1/2} - 2x^{-1/2} \).
02

Separate the Integral

Separate the rewritten integrand into two separate integrals: \( \int_{1}^{9} \left(3x^{1/2}\right) \, dx - \int_{1}^{9} \left(2x^{-1/2}\right) \, dx \). This allows us to integrate each term individually.
03

Integrate Each Term

For the first term \( 3x^{1/2} \), apply the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here, \( n = 1/2 \), so \( \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \). Multiply by 3 to get \( 2x^{3/2} \).For the second term \( 2x^{-1/2} \), again use the power rule. Here, \( n = -1/2 \), so \( \int x^{-1/2} \, dx = 2x^{1/2} \). Multiply by 2 to get \( 4x^{1/2} \).
04

Evaluate Each Integral with Limits

Evaluate the definite integrals from 1 to 9 for both terms:1. For \( 2x^{3/2} \), calculate \( 2 \times \left[ \left( \frac{9^{3/2}}{3/2} \right) - \left( \frac{1^{3/2}}{3/2} \right) \right] = 2 \times (18 - \frac{2}{3}) = 34 \).2. For \( 4x^{1/2} \), calculate \( 4 \times \left[ 9^{1/2} - 1^{1/2} \right] = 4 \times (3 - 1) = 8 \).
05

Subtract to Find the Final Answer

Combine the results of the two integrals by subtracting the second result from the first, reflecting the separation of integrals: \( 34 - 8 = 26 \).

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Key Concepts

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

Power Rule for Integration
The power rule for integration is a fundamental technique used in calculus to find the integral of expressions involving powers of variables. This rule is similar to the power rule for differentiation, but reversed. If you can master this rule, integrating basic power functions becomes quite straightforward.

To apply the power rule for integration, remember the formula:
  • \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \),
where \( n eq -1 \) and \( C \) is the constant of integration.
This tells us that to integrate a term like \( x^n \), you should:
  • Add one to the exponent \( n \).
  • Divide by the new exponent.
For example, for \( 3x^{1/2} \), we apply the rule and find \( 2x^{3/2} \) because when \( n = 1/2 \), the integral becomes \( \frac{x^{3/2}}{3/2} \), and multiplied by 3 gives \( 2x^{3/2} \). Similarly, for \( 2x^{-1/2} \), the integral results in \( 4x^{1/2} \) after applying the power rule and multiplying by 2.

The key takeaway is to always adjust the exponent, divide by the new exponent, and remember the constant of integration for indefinite integrals.
Evaluating Definite Integrals
Definite integrals allow us to find the net area under a curve within specified boundaries—we call these boundaries the limits of integration. The notation for definite integrals includes these limits, written as \( \int_{a}^{b} \), where \( a \) is the lower limit and \( b \) is the upper limit.

To evaluate a definite integral, follow these steps:
  • First, find the antiderivative of the function. This involves integrating as if there were no bounds, using techniques like the power rule.
  • Next, apply the limits of integration. This involves substituting the upper limit and then the lower limit into the antiderivative.
  • Finally, subtract the result of the lower limit substitution from that of the upper limit. This calculation gives you the definitive value of the integral, representing the net area between the function and the x-axis over the interval \([a, b]\).
In our case, integrate \( 2x^{3/2} \) and \( 4x^{1/2} \) separately, substitute the boundaries \( x=9 \) and \( x=1 \), and then perform subtraction to combine the results. This combination of the two calculations is often seen in step-by-step calculus solutions, highlighting the importance of methodically applying limits.
Integral Calculus
Integral calculus is a major area of calculus that deals with integrals and the concept of accumulation. It's integral to understanding changes across various disciplines, from areas under curves to physical quantities like distance and volume.

The fundamental principle behind integral calculus is the idea of finding integrals, either indefinite or definite:
  • Indefinite Integrals: These are general forms of antiderivatives, providing a broad expression for a family of functions without specific evaluation limits.
  • Definite Integrals: These represent a numerical value defined by a specific range, or interval, and are often used to calculate areas under curves or between two points.
Integral calculus employs several rules and techniques, with the power rule being one of the simplest yet powerful tools at a student's disposal. These methods allow integration of different forms of functions, transforming our understanding of complex dynamic systems.

In practice, integral calculus extends to solving real-world problems: from engineering and physics calculations for determining work done by forces, to economics for determining consumer and producer surplus. Understanding the implementation and computation of integrals, especially the use of definite integrals, unlocks a wealth of practical applications.

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Most popular questions from this chapter

Economists use a cumulative distribution called a Lorenz curve to describe the distribution of income between house- holds in a given country. Typically, a Lorenz curve is defined on \([0,1]\) with endpoints \((0,0)\) and \((1,1),\) and is continuous, increasing, and concave upward. The points on this curve are determined by ranking all households by income and then computing the percentage of households whose income is less than or equal to a given percentage of the total income of the country. For example, the point (a/100, b/ 100 ) is on the Lorenz curve if the bottom a \(\%\) of the households receive less than or equal to \(b \%\) of the total income. Absolute equality of income distribution would occur if the bottom a \(\%\) of the households receive a \(\%\) of the income, in which case the Lorenz curve would be the line \(y=x\) . The area between the Lorenz curve and the line \(y=x\) measures how much the income distribution differs from absolute equality. The coefficient of inequality is the ratio of the area between the Lorenz curve and the line \(y=x\) to the area under \(y=x\) . $$\begin{array}{c}{\text { (a) Show that the coefficient of inequality is twice the area }} \\ {\text { between the Lorenz curve and the line } y=x \text { , that is, }} \\ {\text { show that }} \\ {\quad \text { coefficient of inequality }=2 \int_{0}^{1}[x-L(x)] d x}\end{array}$$ $$\begin{array}{c}{\text { (b) The income distribution for a certain country is repre- }} \\ {\text { sented by the Lorenz curve defined by the equation }} \\ {L(x)=\frac{5}{12} x^{2}+\frac{7}{12} x}\end{array}$$ What is the percentage of total income received by the bottom 50\(\%\) of the households? Find the coefficient of inequality.

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