Problem 58 - Calculus

Question

Prove that \(\ln x \leqslant 2(\sqrt{x}-1)\) for \(x>1 .\) Compare the integrats of \(1 / t\) and \(1 / \sqrt{t},\) from 1 to \(x\)

Step-by-Step Solution

Answer

\(\ln x \leq 2(\sqrt{x}-1)\) for \(x>1\) is proven by comparing integrals.

Step 1: Understanding the Integral Comparison

We need to compare the integrals of two functions: \(f(t) = \frac{1}{t}\) and \(g(t) = \frac{1}{\sqrt{t}}\) from 1 to \(x\). To do this, we will evaluate each integral separately.

Step 2: Evaluating the First Integral

Calculate \(\int_{1}^{x} \frac{1}{t} \, dt\), which gives us \( [\ln t]_{1}^{x} = \ln x - \ln 1 = \ln x\).

Step 3: Evaluating the Second Integral

Calculate \(\int_{1}^{x} \frac{1}{\sqrt{t}} \, dt\). First, rewrite \(\frac{1}{\sqrt{t}} = t^{-1/2}\). The antiderivative of \(t^{-1/2}\) is \(2t^{1/2}\). So, \([2\sqrt{t}]_{1}^{x} = 2\sqrt{x} - 2\sqrt{1} = 2\sqrt{x} - 2\).

Step 4: Comparing the Integrals

We now compare \(\ln x\) and \(2\sqrt{x} - 2\). We need to show that \(\ln x \leq 2\sqrt{x} - 2\).

Step 5: Proving the Inequality

The inequality \(\ln x \leq 2(\sqrt{x} - 1)\) holds for all \(x > 1\) since \(\int_{1}^{x} \frac{1}{t} \, dt\) (\(= \ln x\)) is less than \(\int_{1}^{x} \frac{1}{\sqrt{t}} \, dt\) (\(= 2\sqrt{x} - 2\)). This is because \(\frac{1}{t}\) decreases faster than \(\frac{1}{\sqrt{t}}\) in the interval \([1, x]\).

Step 6: Conclusion

Thus proving \(\ln x \leq 2(\sqrt{x} - 1)\) for \(x > 1\) by comparing the two integrals. The inequality holds because the integral of \(\frac{1}{t}\) is smaller than that of \(\frac{1}{\sqrt{t}}\) over the interval from 1 to \(x\).

Key Concepts

Integral Comparison

When dealing with calculus problems, one effective technique is the integral comparison method. It involves evaluating and comparing integrals of different functions over the same interval. In our exercise, we compare the integrals of the functions:

• \( f(t) = \frac{1}{t} \)
• \( g(t) = \frac{1}{\sqrt{t}} \)

from 1 to \(x\). Calculating these integrals helps us understand the behavior of the two functions across the interval. Integrals offer a way to measure the area under the curve, and comparing them tells us how two functions relate to each other over a certain range. This becomes particularly useful in establishing inequalities between different expressions. In this case, it helps us prove that \( \ln x \leq 2(\sqrt{x} - 1) \) for \( x > 1 \). Understanding which function dominates or grows faster can provide insights into solving such inequalities.

Logarithmic Function

The logarithmic function \( \ln x \) is fundamental in calculus due to its unique properties. For positive values of \(x\), it represents the natural logarithm, which is the inverse of the exponential function \( e^x \).
The function is characterized by its slow growth; as \(x\) increases, the value of \( \ln x \) increases at a decreasing rate. This slow rate of change significantly impacts comparisons with other functions, particularly in exercises involving inequalities.
In this problem, the slow growth of the logarithmic function is a key element when comparing it to \( 2(\sqrt{x} - 1) \). Understanding that \( \ln x \) grows slower than \( \sqrt{x} \) over the given interval provides an intuitive foundation for why the inequality \( \ln x \leq 2(\sqrt{x} - 1) \) holds true.

Inequalities in Calculus

Inequalities are essential tools in calculus to infer behaviors of functions or establish bounds. They give us a way to compare values and functions without needing exact solutions. Generally, they leverage known properties or comparisons between functions to draw conclusions.
In math, proving inequalities often involves demonstrating one expression is consistently greater, smaller, or equal to another. In our example, the goal was to show \( \ln x \leq 2(\sqrt{x} - 1) \) for \( x > 1 \). Utilizing integral comparison helps achieve this.
This method allows us to use the nature of integrals: since \( \int_1^x \frac{1}{t} \, dt = \ln x \) and \( \int_1^x \frac{1}{\sqrt{t}} \, dt = 2\sqrt{x} - 2 \), and observing the fundamental behavior of the integrand \( \frac{1}{t} < \frac{1}{\sqrt{t}} \) suggests that accumulating values under this curve yields a lesser result, explaining the inequality.

Antiderivatives

An antiderivative is a function whose derivative gives the original function. It plays a pivotal role in integral calculus by helping us evaluate definite integrals. Finding the antiderivatives allows us to compute the area under a curve or, more generally, to solve integrals.
For instance, in the given problem:

• The antiderivative of \( f(t) = \frac{1}{t} \) is \( \ln t \).
• Similarly, the antiderivative of \( g(t) = \frac{1}{\sqrt{t}} \) is \( 2\sqrt{t} \).

Why does this matter? Because knowing these antiderivatives enables evaluating the definite integrals and directly comparing them. This comparison feeds back into proving our inequality by shedding light upon the accumulated areas under the functions \( \frac{1}{t} \) and \( \frac{1}{\sqrt{t}} \). Understanding antiderivatives fundamentally aids in processes involving integration and plays a vital role in solving problems of this nature.