Problem 12 - Calculus: An Intuitive and Physical Approach

Question

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

Step-by-Step Solution

Answer

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 1: 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 \[ A = \int_{1}^{10} \frac{1}{x} \, dx \]

Step 2: 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. \[ A = \Big[ \ln|x| \Big]_{1}^{10} = \ln|10| - \ln|1| \]

Step 3: 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.

Key Concepts

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.