Real Analysis: Integrals
Integration Techniques
$u$-Substitution
The chain rule for the composition of two differentiable functions $f$ and $g$ states that
$$\dfrac{d}{dx}(f \circ g) = f'(g(x))g'(x).$$
In eligible integrands, the $g(x)$ and $g'(x)$ terms in an integrand may be replaced by $u$ and $u'$ variables to convert the integral from a more complicated function of $x$ to a simpler function of $u,$ which should ideally be easier to integrate. After integration, the antiderivative in terms of $u$ may then be converted back to an antiderivative in terms of $x.$ In more complicated integrals, this process of substitution can happen multiple times, thus requiring multiple re-substitutions after integration.
This process is popularly known as $u$-substitution due to the substitution of complicated terms for simpler ones and the nearly unanimous preference for the letter $u.$
Integration by Parts
The product rule for the derivative of the product of two differentiable functions $f$ and $g$ states that
$$(fg)' = f'g + fg'.$$
By applying the fundamental theorem of calculus to this identity, we can produce the corresponding identity for integrals, known as integration by parts:
$$\int f(x) g'(x) \, dx = f(x)g(x) - \int g(x) f'(x) \, dx.$$
Improper Integration
Definite integrals are defined for closed intervals of the form $[a, b].$ However, these closed intervals can be extended to $\overline{\mathbb{R}}$ to take the forms of $[-\infty, b],$ $[a, \infty],$ and $[-\infty, \infty]$ to allow the calculation of areas under infinite stretches of the number line. The process for doing so involves taking a limit, and the result is called an improper integral, defined for each of the three cases as
$$\begin{array}{rcl}\displaystyle \int\limits_a^{\infty} f(x) \, dx & = & \displaystyle \lim\limits_{b \rightarrow \infty} \int\limits_a^b f(x) \, dx\\ \displaystyle \int\limits_{-\infty}^b f(x) \, dx & = & \displaystyle \lim\limits_{a \rightarrow -\infty}\int\limits_a^b f(x) \, dx\\ \displaystyle \int\limits_{-\infty}^{\infty} f(x) \, dx & = & \displaystyle \lim\limits_{a \rightarrow -\infty} \lim\limits_{b \rightarrow \infty} \int\limits_a^b f(x) \, dx \end{array} $$
Reverse Integration
Integrals are defined in terms of closed intervals which are broken down into infinitely many, infinitesimally small sub-intervals. However, the areas of these rectangles area not always non-negative like they are for actual geometric shapes. If the function is negative in a particular region, then the little rectangles under the curve will have a "negative area." If we can so easily imagine negative area resulting from the height of a little rectangle being negative, we should have no problem at all imagining it as a result of negative width instead.
The length of a closed interval $[a, b]$ is $b - a.$ If one imagines the integral naturally "starting" at $a$ and "ending" at $b,$ then a reversal of the direction, by "starting" at $b$ and "ending" at $a,$ should produce an area of opposite sign. Such a trick can be achieved by $u$-substitution, giving us the following result:
$$ \int\limits_a^b f(x) \, dx = -\int\limits_b^a f(x) \, dx. $$
Symmetric Integrals
Two special cases arise in the evaluation of integrals of even and odd functions over intervals centered around the origin. If $f$ is even, then
$$\int\limits_{-c}^c f(x) \, dx = 2\int\limits_0^c f(x) \, dx.$$
Likewise, if $f$ is instead odd, then
$$\int\limits_{-c}^c f(x) \, dx = 0.$$
Problems
$u$-Substitution: Show that
$$\int\limits_a^b f'(g(x))g'(x) \, dx = \int\limits_{g(a)}^{g(b)} f'(u) \, du.$$
By the chain rule, we see that
$$\dfrac{d}{dx}f(g(x)) = f'(g(x))g'(x).$$
Therefore, by the fundamental theorem of calculus,
$$\int\limits_a^b f'(g(x))g'(x) \, dx = f(g(b)) - f(g(a)).$$
By the same theorem,
$$\int\limits_{g(a)}^{g(b)} f'(u) \, du = f(g(b)) - f(g(a)).$$
Therefore
$$\int\limits_a^b f'(g(x))g'(x) \, dx = \int\limits_{g(a)}^{g(b)} f'(u) \, du.$$
Integration by Parts: Show that if $f$ and $g$ are differentiable on $[a, b],$ then
$$\int\limits_a^b f(x)g'(x) \,dx = f(b)g(b) - f(a)g(a) - \int\limits_a^b f'(x)g(x) \, dx.$$
Let $h(x) = f(x)g(x).$ Then by the product rule,
$$h'(x) = f'(x)g(x) + f(x)g'(x).$$
By the fundamental theorem of calculus,
$$\int\limits_a^b h'(x) \, dx = h(b) - h(a).$$
Expanding $h$ shows that
$$\int\limits_a^b f'(x)g(x) + f(x)g'(x) \, dx = f(b)g(b) - f(a)g(a).$$
By additivity, we see that
$$\int\limits_a^b f'(x)g(x) \, dx + \int\limits_a^b f(x)g'(x) \, dx = f(b)g(b) - f(a)g(a).$$
Finally, a little subtraction gives us the desired form of the result:
$$\int\limits_a^b f(x)g'(x) \, dx = f(b)g(b) - f(a)g(a) - \int\limits_a^b f'(x)g(x) \, dx.$$
Integration by Parts: Show that if $f$ and $f$ are differentiable, then
$$\int f(x) g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx.$$
Note: This proof is very similar to the prior one.
By the chain rule, we see that
$$\dfrac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x).$$
By the fundamental theorem of calculus, it follows that
$$f(x)g(x) = \int f'(x)g(x) + f(x)g'(x) \, dx.$$
Rearranging gives us the desired result:
$$\int f(x)g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx.$$
Reverse integration: Show that $\displaystyle \int\limits_a^b f(x) \, dx = -\int\limits_b^a f(x) \, dx.$
We use $u$-substitution. Let $u(x) = b - x + a.$ Then
$$u(b) = a \text{ and } u(a) = b.$$
Next, note that
$$\dfrac{du}{dx} = -1,$$
so
$$du = -dx.$$
Now substitute into the $u$-substitution formula:
$$ \int\limits_a^b f(x) \, dx = \int\limits_{u(a)}^{u(b)} -f(u) \, du\\ \int\limits_a^b f(x) \, dx = \int\limits_{b}^{a} -f(u) \, du \\ \int\limits_a^b f(x) \, dx = -\int\limits_{b}^{a} f(u) \, du $$
Since $u$ is as much a variable of integration as is $x,$ we can write out the formula in its more recognizable form:
$$\int\limits_b^a f(x) \, dx = -\int\limits_{a}^{b} f(x) \, dx$$
Show that if $f$ is even, then
$$\int\limits_{-c}^c f(x) \, dx = 2\int\limits_0^c f(x) \, dx.$$
Note that
$$ \int\limits_{-c}^c f(x) \, dx = \int\limits_{-c}^0 f(x) \, dx + \int\limits_0^c f(x) \, dx. $$
Next, let $u(x) = -x$ such that $du = -dx.$ Then
$$ \int\limits_{-c}^0 f(x) \, dx = \int\limits_{c}^0 -f(-u) \, du. $$
Applying reverse integration shows that
$$ \int\limits_{c}^0 -f(-u) \, du = \int\limits_0^c f(-u) \, du. $$
Because $f$ is even, we see that $f(-x) = f(x).$ Therefore
$$\int\limits_0^c f(-u) \, du = \int\limits_0^c f(u) \, du.$$
Substituting this back into the original equation gives us
$$ \int\limits_{-c}^c f(x) \, dx = \int\limits_{0}^c f(u) \, du + \int\limits_0^c f(x) \, dx, $$
which simplifies to
$$ \int\limits_{-c}^c f(x) \, dx = 2\int\limits_0^c f(x) \, dx $$
Show that if $f$ is odd, then
$$\int\limits_{-c}^c f(x) \, dx = 0.$$
Note that
$$ \int\limits_{-c}^c f(x) \, dx = \int\limits_{-c}^0 f(x) \, dx + \int\limits_0^c f(x) \, dx. $$
Next, let $u(x) = -x$ such that $du = -dx.$ Then
$$ \int\limits_{-c}^0 f(x) \, dx = \int\limits_{c}^0 -f(-u) \, du. $$
Applying reverse integration shows that
$$ \int\limits_{c}^0 -f(-u) \, du = \int\limits_0^c f(-u) \, du. $$
Because $f$ is odd, we see that $f(-x) = -f(x).$ Therefore
$$\int\limits_0^c f(-u) \, du = \int\limits_0^c -f(u) \, du.$$
Substituting this back into the original equation gives us
$$ \int\limits_{-c}^c f(x) \, dx = -\int\limits_{0}^c f(u) \, du + \int\limits_0^c f(x) \, dx, $$
which simplifies to
$$ \int\limits_{-c}^c f(x) \, dx = 0. $$