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Chapter 12: Problem 12

Find the area between the curve \(y=1 / x\), the \(x\)-axis and the ordinates at \(x=1\) and \(x=10\). Ans. \(2.3026\).

Short Answer

Expert verified
The area between the curve \(y=1 / x\), the \(x\)-axis, and the ordinates at \(x=1\) and \(x=10\) is approximately 2.3026.

Step by step solution

01

Define the Integral

To find the area under the curve, we need to set up an integral of the function from the lower x-bound to the upper x-bound. In this case, the bounds are between x = 1 and x = 10. We can express this integral as \[ A = \int_{1}^{10} \frac{1}{x} \, dx \]
02

Evaluate the Integral

The integral of \(\frac{1}{x}\) is a well-known integral, and its antiderivative is \(\ln|x|\). Apply the fundamental theorem of calculus to evaluate the definite integral. \[ A = \Big[ \ln|x| \Big]_{1}^{10} = \ln|10| - \ln|1| \]
03

Calculate the Numerical Value

Since the natural logarithm of 1 is 0, we simplify the expression to \[ A = \ln|10| - 0 = \ln(10) \] Now, we use a calculator to find the numerical value of \(\ln(10)\), which is approximately 2.3026.

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Key Concepts

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

Area Under a Curve
Understanding the area under a curve is fundamental in calculus, especially when it comes to applications in physics and engineering. It represents the accumulation of a quantity, like distance over time for a changing speed. In the exercise provided, the objective is to find the area between the curve y=1/x, the x-axis, and specific boundaries along the x-axis (ordinates at x=1 and x=10).

To visualize this, imagine plotting the graph of y=1/x and shading the region under this curve from x=1 to x=10. This shaded region represents the area we want to calculate. The definite integral is the tool that allows us to do this effectively, giving us not just the shape of the area but also its exact size.
Fundamental Theorem of Calculus
The fundamental theorem of calculus is the bridge that connects differentiation and integration. It states that if a function f is continuous on a closed interval [a, b], and F is an antiderivative of f on [a, b], then
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
In simpler terms, it allows us to evaluate the definite integral (the area under the curve) between two points on a function's graph by finding the antiderivative. The exercise given is a classic application of this theorem. By finding the antiderivative of the function y=1/x, we are able to use the theorem to calculate the definite integral from x=1 to x=10, thus giving us the sought-after area.
Antiderivative
An antiderivative of a function f(x) is a function F(x) whose derivative is f(x). In other words, F'(x) = f(x). When we take the antiderivative of f(x), we essentially find all possible functions F(x) that could have f(x) as their derivative.

In the context of our exercise, the function f(x) = 1/x has an antiderivative F(x) = ln|x|, which we use to evaluate the integral. Remember, when finding the antiderivative of a function, we also need to include a constant of integration; however, in the case of definite integrals, these constants cancel out, as seen in the calculation of the area.
Natural Logarithm
The natural logarithm, denoted as ln(x), is the logarithm to the base e, where e is an irrational and transcendental number approximately equal to 2.71828. This function is the inverse of the exponential function e^x, meaning that if y = ln(x), then e^y = x.

Logarithms are used to solve equations where the unknown variable is an exponent, and they play a crucial role in the solution of our exercise. The natural logarithm is the antiderivative of 1/x, and this is why it appears in the evaluation of the integral. Calculating ln(10) allows us to find the precise area under the curve between our specified boundaries.

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Most popular questions from this chapter

Find the integral when the derived function has the following values: (a) \(d y / d x=e^{-x} . \quad\) Ans. \(y=-e^{-x}+C\). (b) \(d y / d x=x e^{x^{2}}\) (c) \(d y / d x=e^{\sin x} \cos x\). Ans. \(y=e^{\sin x}+C\). (d) \(d y / d x=\frac{e^{-1 / x}}{x^{2}}\). (e) \(f^{\prime}(x)=e^{-x / 2}\).

The more general statement of how an object loses heat, known as Newton's law of cooling, is that the rate of decrease of temperature of an object is continuous and proportional to the difference between the temperature of the object and that of the surrounding medium. Suppose that the surrounding medium is so vast that the heat it absorbs from the object does not appreciably change the medium's temperature so that it remains constant. Suppose that the object is initially at a temperature of \(100^{\circ}\) and loses temperature continuously at the rate of \(1 / 100\) per minute of the difference between its temperature at time \(t\) and the constant temperature \(T_{0}\) of the surrounding medium. Derive the formula that relates the temperature of the object and time. Ans. \(T-T_{0}=\left(100-T_{0}\right) e^{-0.01 t}\).

Look up in the table of natural logarithms the following quantities: (a) \(\log 3 . \quad\) Ans. \(1.0986 .\) (b) \(\log 5 .\) (c) \(\log 10 . \quad\) Ans. \(2.3026\) (d) \(\log 0.1\). (e) \(\log 0.5 .\) Ans. \(-0.6931\).

In a certain chain of nuclear reactions that take place in a nuclear reactor plutonium decays to uranium 235 and the uranium decays to thorium. The amount of uranium derived from plutonium at any time \(t\) is given by \(u=P\left(1-e^{-\lambda t}\right)\) where \(P\) is the original amount of plutonium. The amount of uranium that has decayed at any time \(t\) is given by \(U=u\left(1-e^{-\lambda t}\right)\). Let us assume that \(P=3400\) grams and that \(\lambda=\frac{1}{10} \log 2\). If the amount of uranium present at any time exceeds what is called the critical mass, namely 800 grams, the reactor will explode. Is the reactor safe? Ans. Disaster.

Find \(y\) when \(y^{\prime}\) has the following values: (a) \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\). Ans. \(y=\log \left(e^{x}+e^{-x}\right)+C\). (b) \(\frac{e^{2 x}-1}{e^{2 x}+1}\) (c) \(\frac{e^{2 x}}{1+e^{2 x}} \quad\) Ans. \(y=\frac{1}{2} \log \left(1+e^{2 x}\right)+C\).
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