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Chapter 8: Problem 6

Find the indefinite integral. $$ \int x^{2} \ln x d x $$

Short Answer

Expert verified
\( \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \)

Step by step solution

01

Identify the Integration Method

The integral \( \int x^2 \ln x \, dx \) involves a product of a polynomial function and a logarithmic function. This is a candidate for integration by parts, where one function will be differentiated and the other integrated.
02

Choose Functions for Integration by Parts

For integration by parts, choose \( u = \ln x \) and \( dv = x^2 \, dx \). This means we will differentiate \( u \) and integrate \( dv \).
03

Differentiate and Integrate

Differentiate \( u = \ln x \) to get \( du = \frac{1}{x} \, dx \). Integrate \( dv = x^2 \, dx \) to get \( v = \frac{x^3}{3} \).
04

Apply the Integration by Parts Formula

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), substitute in your expressions: \[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \left( \frac{x^3}{3} \right) \left( \frac{1}{x} \right) \, dx \].
05

Simplify the Remaining Integral

Simplify the integral \( \int \frac{x^3}{3x} \, dx \) to \( \frac{1}{3} \int x^2 \, dx \).
06

Integrate the Simplified Integral

The integral \( \frac{1}{3} \int x^2 \, dx \) can be computed as \( \frac{1}{3} \times \frac{x^3}{3} + C = \frac{x^3}{9} + C \).
07

Write Down the Final Result

Substitute back the simplified integral into the expression obtained from integration by parts:\[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a technique used in calculus to find the integral of a product of two functions. It's particularly useful when direct integration is not straightforward. The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] Here's how it works:
  • Choose one function to differentiate (\(u\)) and another to integrate (\(dv\)).
  • Calculate the derivative \(du\) and the integral \(v\).
  • Substitute into the formula to simplify the problem.
In the given exercise, we have \( \int x^2 \ln x \, dx \) where choosing \( u = \ln x \) and \( dv = x^2 \, dx \) leads to an easier integral to solve after applying integration by parts. Break down your choice carefully as the method's success heavily relies on this step.
Logarithmic Function
Logarithmic functions, like \( \ln x \), are the inverse of exponential functions. They only take positive real numbers as input and are non-decreasing:
  • The function \( \ln x \) represents the natural logarithm, which is the logarithm to the base \( e \).
  • The derivative of \( \ln x \) is \( \frac{1}{x} \), a fact used in differentiation during integration by parts.
These properties make the logarithmic function a good choice for \( u \) when using integration by parts. Differentiating it is straightforward, reducing complexity in calculations. Remember, the behavior of logarithmic functions is different compared to polynomials, which makes understanding \( \ln x \)'s unique properties crucial in solving integrals.
Polynomial Function
Polynomial functions, such as \( x^2 \), are composed of variables with non-negative integer exponents. They're smooth and continuous, and their derivatives are straightforward to compute:
  • The basic rule for integrating a polynomial \( x^n \) is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
  • In our problem, we integrate \( x^2 \) yielding \( \frac{x^3}{3} \).
This makes polynomials a good candidate for \( dv \) in integration by parts because they are easy to integrate. In the exercise, the polynomial \( x^2 \) when integrated becomes \( \frac{x^3}{3} \). Understanding polynomials' straightforward behavior simplifies calculations, allowing for efficient problem-solving when combined with other functions.

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Most popular questions from this chapter

Evaluate the integral. \(\int \cot x \csc ^{-2} x d x\)

a. Show that \(\int_{0}^{1} 1 / x^{p} d x\) converges if \(p<1\) and diverges otherwise. b. Show that \(\int_{1}^{\infty} 1 / x^{p} d x\) converges if \(p>1\) and diverges otherwise. c. Conclude from (a) and (b) that \(\int_{0}^{\infty} 1 / x^{p} d x\) diverges for every \(p\).

Let $$ I_{n}=\int_{0}^{\infty} x^{n} e^{-x} d x $$ for any nonnegative integer \(n\). Using integration by parts, show that $$ I_{n}=n I_{n-1} $$ for \(n \geq 1\). From this show that $$ I_{n}=n(n-1)(n-2) \cdots 2 \cdot 1 $$

If \(A\) is the amount of radioactive substance at time 0 , then the amount \(f(t)\) remaining at any time \(t>0\) (measured in years) is given by $$ f(t)=A e^{k t} $$ where \(k\) is a negative constant. The mean life \(M\) of an atom in the substance is given by $$ M=-\frac{1}{A} \int_{0}^{\infty} t k f(t) d t $$ a. Using \(k=-1.24 \times 10^{-4}\), find the mean life of a \(\mathrm{C}^{14}\) atom. b. Using \(k=-4.36 \times 10^{-4}\), find the mean life of a radium atom.

Evaluate the integral. \(\int \sin ^{5} w \cot ^{3} w d w\)
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