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Chapter 7: Problem 13

Evaluate the following integrals. $$\int x^{2} \ln x^{3} d x$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int x^{2} \ln x^{3} dx\). Answer: \(\int x^{2} \ln x^{3} d x = \frac{1}{3}x^{3} \ln x^{3} - \frac{1}{6}x^{6} + C\)

Step by step solution

01

Identify u(x) and v'(x)

We want to find functions u(x) and v'(x) such that their product gives the integrand: \(x^{2} \ln x^{3}\). We choose $$u(x) = \ln x^{3} \Rightarrow u'(x) = \frac{1}{x} \cdot \frac{d}{d x}\big(x^{3}\big) = 3x^2$$ and $$v'(x)=x^{2} \Rightarrow v(x)=\frac{1}{3}x^{3}$$.
02

Apply Integration by Parts formula

The integration by parts formula is given by \(\int u(x)v'(x) dx = u(x)v(x) - \int u'(x)v(x) dx\). Using our chosen functions u(x) and v(x) and their derivatives, we have $$\int x^{2} \ln x^{3} d x = \ln x^{3} \cdot \frac{1}{3}x^{3} - \int 3x^{2} \cdot \frac{1}{3}x^{3} d x$$
03

Simplify and Evaluate the Integral

Simplify the expression by multiplying and integrating: $$\int x^{2} \ln x^{3} d x = \frac{1}{3}x^{3} \ln x^{3} - \int x^{5} d x$$ Now evaluate the integral: $$= \frac{1}{3}x^{3} \ln x^{3} - \frac{1}{6}x^{6} + C$$ where C is the constant of integration. The final answer is: $$\int x^{2} \ln x^{3} d x = \frac{1}{3}x^{3} \ln x^{3} - \frac{1}{6}x^{6} + C$$

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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 fundamental part of calculus that focuses on the concept of integration. It is primarily concerned with finding the total size or value, such as area under a curve, of a quantity. It is the inverse process of differentiation. Where differentiation is about rates of change, integration sums out quantities to find the original amount.

Integration is broadly categorized into two types:
  • Definite Integral: Computes the accumulated quantity over a specific interval.
  • Indefinite Integral: Provides a general form of the antiderivative plus a constant (usually denoted as C).

For example, when you are trying to find the area under the curve of the function, integral calculus helps you understand how to accumulate values to compute this area. It's like taking tiny pieces of a curve and adding them together to find the whole. The exercise above involves using the technique of integration by parts to evaluate an integral in the context of integral calculus.
definite integral
A definite integral is a powerful tool in calculus used to find the exact value of an integral over a defined interval. This concept extends the idea of finding an area under a curve between two points. Unlike an indefinite integral, which results in a family of functions, a definite integral will give a specific numerical value.

With definite integrals, you have limits of integration, typically denoted as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the start and end values of the interval over which you're integrating. This means you're accumulating the function \(f(x)\) from \(x = a\) to \(x = b\).

An important factor to keep in mind is that the result of a definite integral can represent various practical interpretations such as:
  • The total distance traveled by an object
  • The net change in amounts of a quantity
  • The area under a curve
In the given exercise, although a definite integral isn't directly calculated, the practice of evaluating parts of an integrand brings us closer to handling definite integrals in broader contexts.
antiderivative
An antiderivative is a core concept in calculus that represents the reverse process of differentiation. It is a function whose derivative is the original function. Understanding antiderivatives is crucial when performing integration, as the goal of integrating a function is essentially to find its antiderivative.

Let's say you have a function \(f(x)\). A function \(F(x)\) is an antiderivative of \(f(x)\) if \(F'(x) = f(x)\). When integrating, you typically find a general form of the antiderivative, often adding a constant \(C\) to account for the family of antiderivatives, this results in \(\int f(x) \, dx = F(x) + C\).

In the context of the original exercise, the step \(\int x^{5} \, dx = \frac{1}{6}x^{6}\) represents finding the antiderivative of \(x^{5}\). The constant \(C\) is added later because any constant has a derivative of zero and thus does not appear in the derivative of \(F(x)\). This completes the process of returning from a rate of change back to the complete function itself.

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Most popular questions from this chapter

Evaluate the following integrals. $$\int_{0}^{a} x^{x}(\ln x+1) d x, a>0$$

The Eiffel Tower property Let \(R\) be the region between the curves \(y=e^{-\alpha x}\) and \(y=-e^{-\alpha x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c>0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves; see the Guided Project The exponential Eiffel Tower.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\).

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

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 2} \frac{d \theta}{\cos \theta+\sin \theta}$$
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