Problem 52 - Calculus Single Variable

Question

It is known that \(\int_{1}^{e} \ln (x) d x=1\). If \(a\) is a positive constant, what is \(\int_{1}^{e} \ln (a x) d x ?\)

Step-by-Step Solution

Answer

The value of \( \int_{1}^{e} \ln(ax) \, dx = 1 + (\ln a) \).

Step 1: Understand the Problem

We are given the integral \(\int_{1}^{e} \ln(x) \, dx = 1\). We need to find the value of \(\int_{1}^{e} \ln(ax) \, dx \) where \(a\) is a positive constant.

Step 2: Use a Substitution

Let's use a substitution to simplify the given integral. Set \( u = ax \), then \( du = a \, dx \). Therefore, \( dx = \frac{du}{a} \). When \( x = 1 \), \( u = a \), and when \( x = e \), \( u = ae \). Thus, the limits of integration change from 1 to \( e \) to \( a \) to \( ae \).

Step 3: Transform the Integral with Substitution

Substituting into the integral, we get:\[\int_{1}^{e} \ln(ax) \, dx = \int_{a}^{ae} \ln(u) \cdot \frac{1}{a} \, du\]Simplify this to:\[\frac{1}{a} \int_{a}^{ae} \ln(u) \, du\]

Step 4: Use Properties of Integrals

We can use the fact that if \( \int_{1}^{e} \ln(x) \, dx = 1 \), then shifting and scaling properties can be applied. Notice that:\[\int_{1}^{e} \ln(x) \, dx = \left[ x \ln(x) - x \right]_{1}^{e}\]Evaluates as \( \left[ e \ln(e) - e \right] - \left[ 1 \ln(1) - 1 \right] = 1 \).

Step 5: Relate Transformed Integral to Known Result

Since \( \int_{1}^{e} \ln(x) \, dx = 1 \), and from substitution, the integral \( \frac{1}{a} \int_{a}^{ae} \ln(u) \, du = \int_{1}^{e} \ln(x) \, dx + a \ln(a) - a \).

Step 6: Simplify and Solve for the Desired Integral

Therefore, we can write:\[\frac{1}{a} \left([ae] \ln(ae) - ae - [a] \ln(a) + a\right)\]which simplifies as\[\frac{1}{a} \left((ae \ln(a) + ae \ln(e) - ae - a \ln(a) + a)\right)\]simplifies to \( ae - ae/a + 1 = 1 \).

Key Concepts

Definite Integrals

In calculus, a definite integral represents the signed area under a curve, typically of a function, from one point to another on the x-axis. This involves calculating the accumulation of quantities, which is not just limited to geometric areas.
Definite integrals are expressed in the form \(\int_{a}^{b} f(x) \, dx\). The notation \([a, b]\) specifies the interval over which the integration is performed, where \(a\) is the lower limit and \(b\) is the upper limit.
To evaluate a definite integral, you substitute the upper and lower limits into an antiderivative (a function whose derivative is the integrand) and subtract the lower limit result from the upper limit result. This is known as the Fundamental Theorem of Calculus.

• The expression \([F(b) - F(a)]\) shows the definite integral of \(f(x)\) from \(a\) to \(b\), where \(F(x)\) is an antiderivative of \(f(x)\).
• The result of a definite integral is a number, which represents the total net area between the curve \(f(x)\) and the x-axis over the interval \([a, b]\).

The crucial takeaway is that definite integrals help calculate total accumulation, whether it's area, volume, or other quantities.

Logarithmic Functions

Logarithmic functions, or logs, are inverse functions of exponential functions and express repeated multiplication as addition. The basic form of a logarithm is \(\ln(x)\), which denotes a natural log with base \(e\). Here, \(e\) is an important mathematical constant approximately equal to 2.718.
Logarithms convert multiplicative relationships into additive ones, making them invaluable for simplifying complex calculations.

• For example, \(\ln(ab) = \ln(a) + \ln(b)\), which illustrates how logarithms transform multiplication into addition.
• The property \(\ln(a^b) = b\ln(a)\) signifies how powers become multiplication within logging.
• The natural log of 1 is always 0: \(\ln(1) = 0\).

When dealing with definite integrals involving \(\ln(x)\), the integral gives insights into logarithms' applications in various mathematical and physical problems:- The integral of \(\ln(x)\) itself is widely encountered. For instance, \(\int \ln(x) \, dx = x\ln(x) - x + C\) shows how integration rewinds the differentiation process of logs.
These properties and equations make solving problems involving logarithmic functions more straightforward and intuitive.

Calculus Substitution Method

The substitution method, also known as \(u\)-substitution, is a powerful technique for solving integrals by simplifying the integrand into a more manageable form. This method is akin to performing the reverse of the chain rule for derivatives.
In practice, we use substitution to transform an integral with respect to one variable into another, making integration feasible:

• First, identify a substitution \(u = g(x)\) that transforms the integrand and the differential \(dx\) into new expressions in terms of \(du\).
• This substitution also changes the integration limits if it's a definite integral. For \(x\) ranging from \(a\) to \(b\), the new limits for \(u\) are \(g(a)\) and \(g(b)\).
• Convert the entire integral into terms of \(u\), and integrate with respect to \(u\).
• Once solved, substitute back to the original variable.

In the context of the problem at hand, we used \(u = ax\), transforming it to \( du = a \, dx\) and realizing \(dx = \frac{du}{a}\).
This allowed us to express the original integral \(\int_{1}^{e} \ln(ax) \, dx\) as \(\frac{1}{a} \int_{a}^{ae} \ln(u) \, du\). Such transformations often reveal deeper relationships and simplify the integration process considerably, especially when dealing with logarithmic or complex functions.