Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume x > 0 and y > 0.


e. The area under the curve y = 1/x and the x-axis on the interval [1, e] is 1.

Verified step by step guidance

1. Recall that the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = e \) is given by the definite integral \( \int_1^e \frac{1}{x} \, dx \).
2. Set up the integral: \( \int_1^e \frac{1}{x} \, dx \). This integral represents the area under the curve between the given limits.
3. Use the fact that the integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| + C \). So, \( \int \frac{1}{x} \, dx = \ln|x| + C \).
4. Evaluate the definite integral by applying the Fundamental Theorem of Calculus: \( \int_1^e \frac{1}{x} \, dx = \left[ \ln x \right]_1^e = \ln e - \ln 1 \).
5. Since \( \ln e = 1 \) and \( \ln 1 = 0 \), the area under the curve on the interval \([1, e]\) is \( 1 - 0 = 1 \). Therefore, the statement is true.

Key Concepts

Definite Integral as Area Under a Curve

The definite integral of a function over an interval represents the net area between the curve and the x-axis within that interval. For positive functions, this integral equals the actual area under the curve. Understanding this helps interpret the integral of y = 1/x from 1 to e as the area under that curve.

Integral of 1/x and the Natural Logarithm

The integral of 1/x with respect to x is the natural logarithm function, ln|x| + C. This means that the definite integral from 1 to e of 1/x dx equals ln(e) - ln(1), which simplifies to 1 - 0 = 1. Recognizing this connection is key to evaluating the area under y = 1/x.

Properties of the Number e

The number e is the base of the natural logarithm, approximately 2.718, and satisfies ln(e) = 1. This property is essential when evaluating integrals involving 1/x, as it directly determines the value of the definite integral from 1 to e, confirming the area under the curve.