Problem 81 - Thomas Calculus

Question

Show that for any number \(a>1\) $$ \int_{1}^{a} \ln x d x+\int_{0}^{\ln a} e^{y} d y=a \ln a $$ (See accompanying figure.) Graph cannot copy

Step-by-Step Solution

Answer

The integrals sum to \( a \ln a \).

Step 1: Evaluate the First Integral

Let's solve the integral \( \int_{1}^{a} \ln x \, dx \). We use integration by parts, where we set \( u = \ln x \) and \( dv = dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \).Using the formula for integration by parts, \( \int u \, dv = uv - \int v \, du \), we have:\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C\]Evaluating from 1 to \( a \), we get:\[\left[ x \ln x - x \right]_1^a = (a \ln a - a) - (1 \cdot \ln 1 - 1) = a \ln a - a + 1\]

Step 2: Evaluate the Second Integral

Now, evaluate \( \int_{0}^{\ln a} e^{y} \, dy \). Since the antiderivative of \( e^y \) is itself, we have:\[\int e^y \, dy = e^y + C\]Evaluating from 0 to \( \ln a \), we get:\[\left[ e^y \right]_0^{\ln a} = e^{\ln a} - e^0 = a - 1\]

Step 3: Combine Results

Add the results of the two integrals from Steps 1 and 2:\[\left[ \int_{1}^{a} \ln x \, dx \right] + \left[ \int_{0}^{\ln a} e^{y} \, dy \right] = (a \ln a - a + 1) + (a - 1)\]Simplify the expression:\[a \ln a - a + 1 + a - 1 = a \ln a\]

Step 4: Conclusion

Thus, the final result is verified. We've shown that:\[\int_{1}^{a} \ln x \, dx + \int_{0}^{\ln a} e^y \, dy = a \ln a\]

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They are used to calculate the accumulation of quantities, such as areas under curves, over an interval. When evaluating a definite integral, you are essentially summing up an infinite number of infinitesimally small products over a defined range.To compute a definite integral, two limits, often referred to as the upper and lower limits, are specified. These limits determine the interval over which the function is integrated. The definite integral is expressed in the form:\[ \int_{a}^{b} f(x) \, dx \]Where:

• \( a \) and \( b \) are the lower and upper limits, respectively.
• \( f(x) \) is the function being integrated.

Evaluating a definite integral involves finding the antiderivative of the function. The antiderivative, evaluated at the upper limit minus the lower limit, gives you the precise quantity accumulated over that interval. A unique characteristic of definite integrals is that they produce a number, rather than a function, due to their fixed integration limits.When working with definite integrals in problems like the one given, make sure to correctly apply the limits to reach the final solution accurately.

Natural Logarithm

The natural logarithm is a logarithmic function with the base of Euler's number \( e \), approximately equal to 2.71828. It is denoted by \( \ln x \). The natural logarithm is the inverse of the exponential function when the base is \( e \). This characteristic makes it extremely useful in calculus, especially when dealing with integrations and derivatives.Here are some key properties of the natural logarithm:

• \( \ln e = 1 \), because \( e^1 = e \).
• \( \ln 1 = 0 \), as \( e^0 = 1 \).
• For any product \( x \, y \), \( \ln(xy) = \ln x + \ln y \).
• For any quotient \( x/y \), \( \ln(x/y) = \ln x - \ln y \).

In equations like \( \int \ln x \, dx \), integration by parts is often employed. This is due to the natural logarithm not having a straightforward antiderivative. In the solved problem, the integration by parts method was crucial for evaluating \( \int_{1}^{a} \ln x \, dx \), offering a clear approach to handle the logarithmic integration efficiently.

Exponential Function

The exponential function is one of the most important functions in mathematics. This function is written as \( e^x \), where \( e \) is the base of the natural logarithm. The exponential function is unique due to its property of having the same rate of growth as its own value.Key characteristics of the exponential function include:

• \( e^x \) is always positive and increases without bounds as \( x \) increases.
• \( e^0 = 1 \) because any number to the power of 0 is 1.
• The derivative of \( e^x \) is itself, \( e^x \).
• It is the inverse function of the natural logarithm \( \ln x \).

In integral calculus, finding the antiderivative of the exponential function \( e^x \) is straightforward, as it is its own antiderivative. This property simplifies many integration problems, like in the example \( \int_{0}^{\ln a} e^y \, dy \). In this calculation, knowing that the antiderivative of \( e^y \) is \( e^y \) allows the integral to be solved readily, leading to elegant solutions in exponential-related integrals.