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Evaluate the following integrals. $$\int x^{a} \ln x d x, a \neq-1$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int x^{a} \ln x dx$$. Answer: $$\int x^{a} \ln x d x = \frac{x^{a + 1} \ln x }{a + 1} - \frac{x^{a + 1}}{(a+1)^2} + C$$

Step by step solution

01

Identifying u and dv

In this case, we will choose: $$ u = \ln x, \quad dv = x^a dx $$
02

Finding du and v

Now we need to find the derivatives and integrals of the chosen functions: $$ du = \frac{1}{x} dx, \quad v = \int x^a dx$$ To find v, we need to integrate the power function \(x^a\): $$ v = \frac{x^{a + 1}}{a + 1} $$
03

Applying Integration By Parts formula

Now we have all the necessary information to apply the integration by parts formula: $$ \int x^{a} \ln x dx = \left(\ln x\right)\left(\frac{x^{a + 1}}{a+1} \right) - \int \left( \frac{x^{a + 1}}{a + 1} \right) \left(\frac{1}{x} dx\right) $$
04

Simplifying the integral

We can simplify the integral by canceling out a factor of x in the numerator and denominator: $$ \int x^{a} \ln x dx = \frac{x^{a + 1} \ln x }{a + 1} - \int \frac{x^a dx}{a + 1} $$
05

Solving the remaining integral

The remaining integral can now be evaluated directly, since it involves a power function: $$ \int x^{a} \ln x dx = \frac{x^{a + 1} \ln x }{a + 1} - \frac{x^{a + 1}}{(a+1)^2} + C $$ So the final answer for the integral is: $$ \int x^{a} \ln x d x = \frac{x^{a + 1} \ln x }{a + 1} - \frac{x^{a + 1}}{(a+1)^2} + C $$

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Key Concepts

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

Integral Calculus
Integral calculus is a core branch of calculus focused on the concepts of integration. Integration is essentially the reverse process of differentiation and is used to find areas under curves, accumulated quantities, and other physical interpretations such as work done by a force. When we encounter problems in calculus involving the area underneath a curve or the total accumulation of quantity, we turn to integral calculus.

The integration by parts technique is one of the fundamental tools used in integral calculus, particularly when dealing with the product of two functions. Stemming from the product rule of differentiation, this method allows us to break down complex integrals into simpler parts. As seen in the original exercise, integration by parts involves choosing an appropriate function for differentiation and a function for integration, leading us to the simpler integral via the formula:

\[ \int u \, dv = uv - \int v \, du \]

This formula plays a crucial role when dealing with integrals that are not straightforward, such as \( \int x^a \ln x \, dx \). By strategically selecting our "\( u \)" and "\( dv \)", we can simplify the problem and solve the integral.
  • Choose "\( u \)" to be a function that simplifies when differentiated.
  • Choose "\( dv \)" such that its antiderivative is easy to compute.
Applications of integral calculus extend to physics, engineering, and many fields where accumulation and area are important considerations.
Power Functions
Power functions are expressions where a variable is raised to a constant power, symbolically defined as \( x^a \), where \( a \) is a constant. When dealing with such functions in calculus, particularly in integration, we utilize specific rules to find antiderivatives. The formula \( \int x^a \, dx = \frac{x^{a+1}}{a+1} + C \) applies for all non-zero \( a \) and provides a straightforward way to integrate power functions.

In the exercise, when we identified \( dv = x^a \, dx \), we leveraged our understanding of power functions to compute the antiderivative. The integration of \( x^a \) is crucial for determining the part of the expression "\( v \)" in the integration by parts formula. Knowing the power rule for integration is one aspect that simplifies our calculus work:
  • Increase the exponent by one.
  • Divide by the new exponent.
This rule is highly important in integrals involving polynomial expressions. Mastering the integration of power functions is essential because these forms frequently appear in algebraic expressions that require integration, making them fundamental to understanding other complex calculus operations.
Natural Logarithm
The natural logarithm, expressed as \( \ln x \), is a logarithmic function with a base equal to \( e \approx 2.71828 \), which is the unique number such that the derivative of \( e^x \) with respect to \( x \) is \( e^x \). This logarithmic function commonly appears in calculus problems involving growth and decay, optimization, and other mathematical models.

Logarithmic functions, like the natural logarithm, have interesting properties that make them useful, especially when they appear as parts of integrands, like in the original problem \( \int x^a \ln x \, dx \). When using integration by parts in the solution provided, \( \ln x \) was chosen as \( u \) because its derivative is simple and reduces complexity in calculations.

Some essential properties of the natural logarithm include:
  • The derivative \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).
  • Logarithmic identities simplify expressions and integrals.
  • \( \ln(1) = 0 \) and \( \ln(e) = 1 \).
Understanding these properties aids in manipulating expressions involving \( \ln x \), making it easier to integrate and apply in various mathematical problems that arise in fields like economics, physics, and engineering.

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Most popular questions from this chapter

Trapezoid Rule and Simpson's Rule Consider the following integrals and the given values of \(n .\) a. Find the Trapezoid Rule approximations to the integral using \(n\) and \(2 n\) subintervals. b. Find the Simpson's Rule approximation to the integral using \(2 n\) subintervals. It is easiest to obtain Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(8 .\) c. Compute the absolute errors in the Trapezoid Rule and Simpson's Rule with \(2 n\) subintervals. $$\int_{0}^{\pi / 4} \frac{d x}{1+x^{2}} ; n=64$$

Use numerical methods or a calculator to approximate the following integrals as closely as possible. The exact value of each integral is given. $$\int_{0}^{\pi / 2} \ln (\sin x) d x=\int_{0}^{\pi / 2} \ln (\cos x) d x=-\frac{\pi \ln 2}{2}$$

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}(\) coulombs ) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}\).

Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\). b. Change variables by letting \(y=1 / x\). c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0\). d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\). (Source: Mathematics Magazine \(59,1,\) Feb 1986 )

Determine whether the following integrals converge or diverge. $$\int_{1}^{\infty} \frac{2+\cos x}{\sqrt{x}} d x$$

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