Question

Write an integral to express the area under the graph of \(y=\frac{1}{t}\) from \(t=1\) to \(e^{x}\) and evaluate the integral.

Short answer

The area is equal to \( x \).

Step-by-step solution

Define the Integral

To find the area under the curve of the function \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = e^x \), we need to set up the integral. The integral that represents the area is: \[ \int_{1}^{e^x} \frac{1}{t} \, dt \]

Identify the Antiderivative

The function \( y = \frac{1}{t} \) is the derivative of \( \ln|t| \). Therefore, the antiderivative of \( \frac{1}{t} \) is:\[ \ln|t| \]

Evaluate the Definite Integral

Now, evaluate the integral \( \int_{1}^{e^x} \ln|t| \, dt \). Apply the fundamental theorem of calculus which states:\[ \int_{a}^{b} f(t) \, dt = F(b) - F(a) \]Where \( F(t) \) is the antiderivative of \( f(t) \).Thus, we calculate:\[ F(e^x) = \ln(e^x) = x \] \[ F(1) = \ln(1) = 0 \]

Calculate the Area

Subtract the two evaluations of the antiderivative to find the definite integral's value:\[ \int_{1}^{e^x} \frac{1}{t} \, dt = \ln(e^x) - \ln(1) = x - 0 = x \]

Conclusion

Therefore, the area under the graph of \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = e^x \) is equal to \( x \).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concepts of differentiation and integration, two cornerstone concepts in calculus. When we discuss areas under a curve, such as in definite integrals, this theorem is crucial. It tells us that if we have a continuous function, we can find the accumulated area from one point to another by evaluating its antiderivative at specific points. For our exercise, the function involved is \( y = \frac{1}{t} \), and the antiderivative is \( \ln |t| \). The theorem states:

• If \( F(t) \) is an antiderivative of \( f(t) \), then \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \).
• This allows us to easily transition from finding slopes (through derivatives) to finding areas (through definite integrals).

When we performed this with our specific function, we set the bounds from 1 to \( e^x \). This resulted in finding the area as \( x \), after evaluating \( F(t) \) at both bounds and subtracting them.

Antiderivative of 1/t

Finding the antiderivative is a key step when handling integrals. The function \( \frac{1}{t} \) has a very special antiderivative: the natural logarithm, specifically \( \ln |t| \). Here's why:

• The derivative of \( \ln |t| \) is \( \frac{1}{t} \). This makes it the perfect candidate as the antiderivative.
• Remember, taking the derivative of \( \ln |t| \) gives us a function that describes the steepness or slope of \( \ln |t| \) at any given point \( t \).

Having \( \ln |t| \) as our antiderivative means we can effectively convert our problem of finding the integral and area into evaluating the natural log at two points.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental mathematical function that arises in calculus. It is particularly useful because of its relationship with the exponential function. In terms of the integral of \( \frac{1}{t} \), understanding \( \ln \) is key.

• The natural logarithm \( \ln(t) \) represents the exponent needed to raise \( e \) (approximately 2.718) to obtain \( t \).
• In calculating the integral from \( t=1 \) to \( e^x \), we appealed to properties of logarithms. Specifically, \( \ln(e^x) = x \) simplifies our work substantially, as seen in our exercise.
• It helps simplify evaluations involving the growth rates and time-based problems.

Thus, mastering \( \ln \) not only aids in calculating integrals for functions like \( \frac{1}{t} \) but also enhances one's understanding of growth and change.