Question

Complete the table by identifying \(u\) and \(d u\) for the integral. $$ \int f(g(x)) g^{\prime}(x) d x \quad \underline{u=g(x)} \quad \underline{d u=g^{\prime}(x) d x} $$ $$ \int x^{2} \sqrt{x^{3}+1} d x $$

Short answer

Function \(u\) is \(x^3+1\), and \(du\) is \(3x^2dx\). But we rewrite \(du\) as \(du/3 = x^2 dx\) to match perfectly with the integral.

Step-by-step solution

Identification of the function 'u'

In this integral, look for a function and its derivative. Compare the integral \(\int x^{2} \sqrt{x^{3}+1} dx\) and the standard formula \(\int f(g(x)) g'(x) dx\). Therefore, \(u = x^3 + 1\). This choice of \(u\) simplifies the integral since its derivative \(x^2\) is present in the integral.

Identification of 'du'

Differentiate \(u\) with respect to \(x\) to obtain \(du\). By applying the chain rule, the derivative of \(x^3 + 1\) is 3x^2 dx. This is not exactly present in the integral. However, \(x^2 dx\) is a part of the derivative and therefore, can be written as a part of \(du\). Hence, \(du = 3x^{2} dx\).

Re-write 'du' to match the integral

We see that \(du = 3x^2 dx\) and in the integral, we have \(x^2 dx\), not \(3x^2 dx\). Therefore, we need to adjust \(du\) to match the integral. We can write \(du/3 = x^2 dx\) which now fits perfectly with the integral

Key concepts

Integral Calculus

Integral calculus, one of the two main operations in calculus, is all about finding the total or accumulated value such as areas under curves, volumes, and other quantities where you are adding up an infinitely large number of infinitesimally small data points. Integral calculus is often introduced after differential calculus because it is conceptually easier to start with the slopes of curves and rates of change before thinking about total accumulation.

One of its fundamental principles is the reverse operation of differentiation - if you know how a function is derived, you can figure out its antiderivative, or integral. There's much more to integral calculus than just the anti-derivative part, though. It includes many techniques of integration, and among these techniques is the method of u-substitution, which often simplifies complex integrals by changing variables and makes them easier to solve.

Chain Rule

The chain rule is a formula to compute the derivative of a composite function. In simpler terms, it’s a way to find out how fast a 'function of a function' is changing.

Let’s say you have two functions, one nested inside the other. If you want to know the derivative of this nesting, you have to take the derivative of the outer function, leaving the inner function alone (as if it were just an 'x') and then multiply it by the derivative of the inner function. Symbolically, if we have a function h(x) that is the composition of functions f and g, such as h(x) = f(g(x)), the derivative h'(x) is h'(x) = f'(g(x)) * g'(x).

Understanding the chain rule is crucial for performing u-substitution in integration because it is used to transform the integral into a simpler form by identifying a section of the integrand as a single variable u, and the differential part of it (du) as the derivative of u.

Integration Techniques

Integration techniques refer to various methods used to compute integrals. Different functions and equations often require different integration strategies. Some of the common techniques besides u-substitution include integration by parts, partial fraction decomposition, and trigonometric substitution. Each method has its own place, and knowing when to use which technique is a foundational skill in integral calculus.

U-substitution is particularly useful as it is essentially the application of the chain rule in reverse and is used in situations where integrands involve the composition of two functions. The value of u-substitution is seen in how it transforms complicated integrals into simpler, more manageable forms. As in our exercise, by identifying the inner function g(x) and its derivative g'(x), we cleverly rewrite the integral in terms of u and du, which often leads to an integral that is straightforward to solve. This method not only reduces complexity but also improves understanding by encouraging the identification of patterns in different functions.