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Chapter 5: Problem 71

Area In Exercises \(71-74\) , find the area of the region bounded by the graphs of the equations. Use a graphing utility to verify your result. $$ y=\frac{x^{2}+4}{x}, \quad x=1, \quad x=4, \quad y=0 $$

Short Answer

Expert verified
The area of the region bounded by the given function and lines is \(8 + 4ln4\) square units.

Step by step solution

01

Simplify the function

First, simplify \( y = (x^2 + 4) / x \) to the function \( y = x + 4/x \). This is done by dividing each term in the numerator by x.
02

Set up the integral

Now, set up the definite integral between x = 1 and x = 4 for the function \( y = x + 4/x \). This will give the area of the region bounded by the graphs. The integral will look like this: \( \int_{1}^{4} {x + 4/x} dx \). This represents the signed area between the function and the x axis between x = 1 and x = 4.
03

Evaluate the integral

Next, evaluate \( \int_{1}^{4} {x + 4/x} dx \) using the fundamental theorem of calculus by calculating the antiderivative at 4 and subtracting the antiderivative at 1. The antiderivative of this function is \(0.5*x^2 + 4*ln|x|\). Therefore, we will have \( (0.5*4^2 + 4*ln|4|) - (0.5*1^2 + 4*ln|1|) = 8 + 4ln4 \).
04

Finalize the answer

After calculating, the area of the region bounded by the graphs is \(8 + 4ln4\) square units.
05

Verification

This result can be checked by graphing the function \( y = x + 4/x \) along with the lines x = 1, x = 4, and y = 0. The area of the shaded region between these lines will be the result of the integral from step 4.

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Key Concepts

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

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a bridge between differentiation and integration. It tells us how to evaluate definite integrals, which are used to find areas under curves. This theorem has two parts:
  • The first part guarantees the existence of antiderivatives for continuous functions. It states that if you integrate a function over an interval, the result is related to the values of its antiderivative at the endpoints.
  • The second part provides a way to calculate this area: by finding the antiderivative and evaluating it at the upper and lower limits, then taking the difference.
In our problem, we used this theorem to evaluate the definite integral \( \int_{1}^{4} {x + \frac{4}{x}} \, dx \) by calculating the antiderivative at points 4 and 1 and subtracting. This gave us the area under the curve.
Area Under Curve
Finding the area under a curve is a common problem in calculus. It involves calculating the space between the graph of a function and the x-axis across an interval. This is useful in a variety of real-world applications, like physics and economics.
  • To find this area, we set up a definite integral with the function representing the curve.
  • The limits of integration, in this case, \(x = 1\) and \(x = 4\), define the interval over which we're finding the area.
  • The area calculated is actually a net area, meaning it considers areas below the x-axis as negative.
For the function \(y = x + \frac{4}{x}\), the definite integral from 1 to 4 helps us find the area bounded by the curve and the x-axis, giving us the total area of \(8 + 4\ln 4\).
Antiderivatives
Antiderivatives are functions that reverse differentiation. If you have a function \(f(x)\), an antiderivative \(F(x)\) satisfies the condition \(F'(x) = f(x)\). Antiderivatives are crucial for solving integrals.
  • In our problem, we needed to find the antiderivative of \(y = x + \frac{4}{x}\).
  • Integrating this gives \(0.5x^2 + 4\ln|x|\), which we use to find exact values for the definite integral.
  • These calculations allow us to solve for specific values, such as the definite integral from 1 to 4, thus determining the precise area under the curve.
Antiderivatives simplify the process of integration, turning complex problems into straightforward calculations.

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