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Chapter 13: Problem 8

Evaluate the integrals. $$ \int_{2}^{3}\left(x+\frac{1}{x}\right) d x $$

Short Answer

Expert verified
The short version of the answer is: $$\int_{2}^{3}\left(x+\frac{1}{x}\right) d x = \frac{5}{2} + \ln \frac{3}{2}$$

Step by step solution

01

Find the antiderivative of the function

We begin by finding the antiderivative of the function \(x + \frac{1}{x}\). We can find the antiderivative of each term separately, and then add them together. The antiderivative of \(x\) is: $$ \int x dx = \frac{x^2}{2} + C_1 $$ The antiderivative of \(\frac{1}{x}\) is: $$ \int \frac{1}{x} dx = \ln |x| + C_2 $$ Thus, the antiderivative of the given function is: $$ \int \left(x + \frac{1}{x}\right) dx = \frac{x^2}{2} + \ln |x| + C $$
02

Evaluate the definite integral

Now we will evaluate this antiderivative for the given interval \([2, 3]\). To do this, we apply the Fundamental Theorem of Calculus: $$ \int_{2}^{3} \left(x + \frac{1}{x}\right) dx = \left(\frac{x^2}{2} + \ln |x|\right) \Big|_{2}^{3} $$ Now we evaluate the antiderivative at the upper and lower bounds of the interval, and subtract the lower bound result from the upper bound result: $$ \begin{aligned} \left(\frac{x^2}{2} + \ln |x|\right)\Big|_{2}^{3} &= \left(\frac{3^2}{2} + \ln |3|\right) - \left(\frac{2^2}{2} + \ln |2|\right) \\ &= \left(\frac{9}{2} + \ln 3\right) - \left(2 + \ln 2\right) \end{aligned} $$
03

Simplify and find the final result

Finally, we simplify the expression: $$ \begin{aligned} \left(\frac{9}{2} + \ln 3\right) - \left(2 + \ln 2\right) &= \frac{9}{2} - 2 + \ln 3 - \ln 2 \\ &= \frac{5}{2} + \ln \frac{3}{2} \end{aligned} $$ Therefore, the value of the definite integral is: $$ \int_{2}^{3}\left(x+\frac{1}{x}\right) d x = \frac{5}{2} + \ln \frac{3}{2} $$

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Key Concepts

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

Antiderivatives
Antiderivatives are a fundamental part of calculus, often referred to as the reverse process of differentiation. If you have a function and want to find its original form before differentiation, you're looking for its antiderivative. In the exercise, the function given is a sum: \(x + \frac{1}{x}\). We find the antiderivative by dealing with each part separately, making things very manageable.
  • For \(x\), the antiderivative is \(\frac{x^2}{2}\), since the derivative of \(\frac{x^2}{2}\) is \(x\).
  • For \(\frac{1}{x}\), the antiderivative is \(\ln |x|\), because differentiating \(\ln |x|\) gives you \(\frac{1}{x}\).
Once you have these, combine them to get the complete antiderivative: \[ \frac{x^2}{2} + \ln |x| + C \]Here, \(C\) is the constant of integration, which we ignore when computing definite integrals.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation with integration, providing a neat way to evaluate definite integrals. It states that if you have a continuous function on an interval and its antiderivative, you can find the definite integral over that interval. In simple terms, this theorem says that integrating a function over a specific interval is the same as finding the change in the antiderivative over that interval.
The exercise applies this by taking the antiderivative \(\frac{x^2}{2} + \ln |x|\) and calculating its values at the endpoints of the interval \([2, 3]\):
  • Evaluate at upper bound (\(x = 3\)): \(\frac{3^2}{2} + \ln 3\)
  • Evaluate at lower bound (\(x = 2\)): \(\frac{2^2}{2} + \ln 2\)
Finally, subtract the result at the lower bound from that at the upper bound.
Integral Evaluation
Evaluating integrals involves solving them over given limits to find out the total accumulation of the function between those limits. For definite integrals, you get a numeric value.After applying the Fundamental Theorem of Calculus in our exercise, you subtract the value of the antiderivative at the lower limit from its value at the upper limit:
\[ \left(\frac{9}{2} + \ln 3\right) - \left(2 + \ln 2\right) \]This simplifies to:
\[ \frac{5}{2} + \ln \frac{3}{2} \]
  • The subtraction requires basic simplification by combining like terms.
  • This process results in a single number, \(\frac{5}{2} + \ln \frac{3}{2}\), which is the total area under the curve from 2 to 3 for the function \(x + \frac{1}{x}\).
This solution provides not just a number, but insight into how areas under curves are calculated in real-world scenarios.

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Most popular questions from this chapter

Calculate the total area of the regions described. Do not count area beneath the \(x\) -axis as negative. HINT [See Example 6.] Bounded by the \(x\) -axis, the curve \(y=x e^{x^{2}}\), and the lines \(x=0\) and \(x=(\ln 2)^{1 / 2}\)

(Compare Exercise 79 in Section 13.2.) The number of research articles in the prominent journal Physical Review written by researchers in Europe can be approximated by \(E(t)=\frac{7 e^{0.2 t}}{5+e^{0.2 t}}\) thousand articles per year \((t \geq 0)\) where \(t\) is time in years \((t=0\) represents 1983\() .^{49}\) Use a definite integral to estimate the number of articles were written by researchers in Europe from 1983 to 2003 . (Round your answer to the nearest 1,000 articles.) HINT [See Example 7 in Section 13.2.]

Calculate the total area of the regions described. Do not count area beneath the \(x\) -axis as negative. HINT [See Example 6.] Bounded by the line \(y=x\), the \(x\) -axis, and the lines \(x=0\) and \(x=1\)

(Compare Exercise 42 in Section 13.3.) The rate of U.S. per capita sales of bottled water for the period 2000-2008 can be approximated by \(s(t)=0.04 t^{2}+1.5 t+17\) gallons per year \(\quad(0 \leq t \leq 8)\) where \(t\) is the time in years since the start of \(2000^{38}\) Use the FTC to estimate the total U.S. per capita sales of bottled water from the start of 2003 to the start of 2008 . (Round your answer to the nearest gallon.)

Evaluate the integrals. $$ \int_{1}^{2} \frac{e^{1 / x}}{x^{2}} d x $$
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