Question

State whether you would use integration by parts to evaluate the integral. If so, identify what you would use for \(u\) and \(d v\). Explain your reasoning. $$ \int x \ln x d x $$

Short answer

Yes, the integral \(\int x \ln x dx\) can be evaluated using integration by parts. The choice for \(u\) and \(dv\) would be \(\ln x\) and \(x dx\) respectively, because their derivative and integral are simpler. The final solution to the integral is \(\frac{1}{2}x^2 \ln x - \frac{1}{4} x^2\).

Step-by-step solution

Organize the Integral

Observe that the integral is a product- \(x\) (which is easily integrable) and \(\ln x\) (its derivative is simpler). The integral is \(\int x \ln x d x\).

Apply Integration by Parts

The form of integration by parts is \(\int udv=uv-\int vdu\). Let one part to differentiate (u) be \(\ln x\), and one part to integrate (dv) be \(x dx\) so that differentiation simplifies the first function and integration won't complicate the second function.

Find du and v

To complete the parts of the theorem, differentiate \(u\) and integrate \(d v\). Find \(du\) by differentiating \(ln x\) (which gives \(du=\frac{1}{x} dx\)), and find \(v\) by integrating \(x dx\) (which gives \(v=\frac{1}{2} x^2\)).

Apply the Formula of Integration by Parts

On applying the formula of integration by parts \(\int udv=uv-\int vdu\) at this point yields: \( \int x \ln x dx = u.v - \int v du\)Substitute the known values for \(u, v, \) and \(du\). This gives \(\frac{1}{2}x^2 \ln x - \int \frac{1}{2} x dx\).

Solve the Integral

The integral \(\int \frac{1}{2} x dx\) can now be easily solvable directly which gives \(\frac{1}{4} x^2\).

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focusing on the processes of integration and its applications. It's part of the core of calculus, alongside differential calculus. Essentially, it's about finding the area under curves and can be thought of as the reverse process of differentiation. When faced with an integral, the primary goal is to simplify the expression until it reaches a form that can be integrated directly.

With functions that are products of simpler functions, as in the exercise \(\int x \ln x dx\), integral calculus provides techniques like integration by parts to break them down into solvable pieces. Understanding that integral calculus not only allows us to solve for areas but also to solve a variety of physical problems, we can see its importance in fields like physics, engineering, economics, and more.

Logarithmic Integration

Logarithmic integration involves integrating functions that contain logarithms. The natural logarithm function, \(\ln x\), appears frequently in calculus due to its unique properties and its derivative, \(\frac{1}{x}\).

When integrating a product of a logarithm with another function – especially when the other function is its derivative – it often simplifies calculations dramatically. In our exercise where \(u = \ln x\), integrating by parts makes perfect sense as the logarithmic function changes into a much simpler algebraic function upon differentiation.

U-Substitution in Integration

U-substitution is a method used to simplify integrals by substituting part of the integral with a new variable, often termed as \( u \). The goal is to transform a complex integral into a simpler one that we can integrate easily.

One commonly looks for a function within the integral whose derivative is also present. However, when the integral is a product of functions such as \(x \ln x\), this substitution might not be straightforward or even necessary. In our exercise, while u-substitution isn't directly used, the concept of choosing a part of the integral (\(u\)) to be differentiated plays a similar role, leading to a simpler integral to work with after applying integration by parts.

Definite Integration Techniques

Definite integration refers to calculating the integral within specific bounds. This provides exact values representing the net area under a curve, between two points. Techniques for definite integration include finding antiderivatives, applying the Fundamental Theorem of Calculus, integration by parts, and u-substitution.

In this exercise with the indefinite integral \(\int x \ln x dx\), we aren't given limits of integration, but the methods we use can be extended to definite integrals. After finding the indefinite integral using integration by parts, if bounds were provided, one would simply evaluate the resulting expression at those bounds to compute the definite integral.