Chapter 16: Problem 10
If \(I_{m}=\int_{1}^{x}(\log x)^{m} d x\) satisfies the relation \(I_{m}=k-l I_{m-1}\), then (A) \(k=e\) (B) \(l=m\) (C) \(k=\frac{1}{e}\) (D) None of these
Short Answer
Expert verified
Correct options are (A) and (B): k=e, l=m.
Step by step solution
01
Differentiate the Given Function
Start by differentiating both sides of the equation for \(I_m\) with respect to \(x\). We have \(I_m = \int_{1}^{x} (\log x)^m \, dx\), so when differentiated, it gives us \(\frac{dI_m}{dx} = (\log x)^m\).
02
Compare with the Differentiation
Given the relation \(I_m = k - l I_{m-1}\), differentiate both sides with respect to \(x\). This gives us \(\frac{dI_m}{dx} = -l \cdot \frac{dI_{m-1}}{dx}\). Now, let's look at \(\frac{dI_m}{dx} = (\log x)^m\) and step directly to \(\frac{dI_{m-1}}{dx} = (\log x)^{m-1}\).
03
Substitute and Simplify
Since \(\frac{dI_m}{dx} = (\log x)^m\) and also \(=-l(\log x)^{m-1}\), equate these expressions: \((\log x)^m = -l(\log x)^{m-1}\). Divide through by \((\log x)^{m-1}\) to solve for \(l\).
04
Solve for 'l'
From \( (\log x)^m = -l(\log x)^{m-1} \), dividing by \((\log x)^{m-1}\), we get \( \log x = -l \). Coefficient comparison implies \( l = m\).
05
Integrate and Solve for 'k'
Integrate \(\int (\log x)^m dx = k - l \int (\log x)^{m-1} dx\) to solve for \(k\). This gives us \(k = I_m + mI_{m-1}\). For \(x=1\), \(I_m\) is a constant and evaluates to 0 at lower limit, so \(k = e\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
In calculus, differentiation involves finding the derivative, a powerful tool that tells us the rate of change of a function. When we differentiate, we are looking for how a quantity changes in relation to another. For example, if you consider the function \( I_m = \int_{1}^{x} (\log x)^m \, dx \), differentiating it with respect to \( x \) allows us to transform this integral into a more manageable expression.
Why is this important?
Why is this important?
- Differentiation helps to provide insightful information about the behavior of functions over their domains.
- It facilitates the comparison between different expressions by breaking them down into simpler forms.
- Understanding the underlying structure of the function makes it easier to solve integrals.
Definite Integrals
A definite integral, such as \( \int_{1}^{x} (\log x)^m \, dx \), is key in solving functions within specified limits. The definite integral calculates the area under a curve between two points, providing a numerical result instead of a function like an indefinite integral.
Why are definite integrals essential?
Why are definite integrals essential?
- They provide the exact accumulated value over an interval, which is crucial in many applications like physics and engineering.
- They translate a function into a definite number which has practical meaning in real-world scenarios.
- They simplify problems focusing on a specific region of interest.
Logarithmic Functions
Logarithmic functions, represented by \( \log x \), are the inverse of exponential functions and widely used in scientific calculations. Developed from the need to simplify multiplication and division into addition and subtraction, logs play a crucial role in various fields.
Key points about logarithmic functions:
Key points about logarithmic functions:
- They transform multiplicative relationships into additive ones. This property is particularly useful when dealing with exponential growth or decay.
- Logarithms form a basis for log scales, which are used in various scientific data representations like the Richter scale.
- They simplify complex calculations and are integral in determining unknown values within equations.