3 Integrals

At the very beginning of this article, we briefly introduced integration, a crucial part of integral calculus, as the summation of infinitesimal differences. When these infinitesimal differences refer to lines or curves, then the product, the summed result is the area under the curve of a given function; when referring to surfaces, the result is a volume; and so forth.

Let us first introduce a definite integral. A definite integral of a function f(x) from a to b is denoted as

    \[\int_{a}^{b} f(x)\,dx\]

provided f is a continuous function defined on x\in [a,b].

A definite integral is commonly referred to as the area under the curve. However, this is only a result, but a more intuitive way of understanding is as a multiplication of infinitesimal difference (dx) with varying, i.e. changing, values of a function.

A classic way of approximating a definite integral is the famous Riemann sum. Riemann sum is computed as the sum of areas partitioned into n rectangles, trapezoids, or other shapes of choice. When this n goes to \infty, the Riemann sum becomes a definite integral.

Let us illustrate the Riemann sum with an example [10].

Example. Find the Riemann sum of a continuous function f(x) on the interval [a,b].

First, we divide the interval into n equally spaced subintervals by letting

    \[\Delta x=\frac{b-a}{n}\]

Then, we have [x_{k-1},x_k], where x_k=a+k\Delta x,
for k=0,1,2,\cdots,n.

Then, we have the approximate k-th strip area of a rectangle with width \Delta x and height f(x_k) (Figure 5 [10]). Therefore, the Riemann sum R_n is

    \begin{align*} R_n &= f(x_1)\Delta x+f(x_2)\Delta x+\cdots+f(x_n)\Delta x\\ &= \Sigma_{k=1}^{n} f(x_k)\Delta x \end{align*}

Now we formally define the the area S under the curve on a given interval [10].

Definition. (Area under Curve)
The area S under the curve on the given interval of a continuous function f(x) is the limit of the Riemann sum. That is,

    \begin{align*} S &= \lim_{n \to \infty} R_n\\ &= \lim_{n \to \infty} \left[ f(x_1)\Delta x+f(x_2)\Delta x+\cdots+f(x_n)\Delta x\right]\\ &= \lim_{n \to \infty} \left[ \Sigma_{k=1}^{n} f(x_k)\Delta x\right] \end{align*}

In fact, we can approximate the Riemann sum with the left-hand side height of a given strip, where L_n denotes the rectangles with a left-hand side height, i.e.

    \begin{align*} S &= \lim_{n \to \infty} L_n\\ &= \lim_{n \to \infty} \left[ f(x_0)\Delta x+f(x_1)\Delta x+\cdots+f(x_{n-1})\Delta x\right]\\ &= \lim_{n \to \infty} \Sigma_{k=0}^{n-1} \left[f(x_{k})\Delta x\right] \end{align*}

Consequently,

Definition. (Definite Integral)
The definite integral of f from a to b is

    \begin{align*} \int_{a}^{b} f(x)dx &= \lim_{n \to \infty} \left[ f(x_0)\Delta x+f(x_1)\Delta x+\cdots+f(x_{n-1})\Delta x\right]\\ &= \lim_{n \to \infty} \left[ f(x_1)\Delta x+f(x_2)\Delta x+\cdots+f(x_n)\Delta x\right]\\ &= \lim_{n \to \infty} \Sigma_{k=0}^{n-1} \left[f(x_{k})\Delta x\right]\\ &= \lim_{n \to \infty} \Sigma_{k=1}^{n} \left[f(x_{k})\Delta x\right] \end{align*}

As opposed to the area under the curve, in fact a \textbf{definite integral}, let us now think of a function F whose derivative f is an integrand, the function to be integrated. The integrated product F is called the antiderivative of f and we express this in the form of an indefinite integral as follows

    \[\int f(x)\,dx=F(x) \text{ i.e. } F'(x)=f(x)\]

Some commonly used indefinite integrals are provided as follows [10].

  • \int k \, dx = kx+C
  • \int x^n \, dx = \frac{x^{n+1}}{n+1}+C, where n\neq-1
  • \int \sec^2{x} \, dx = \tan{x}+C
  • \int \sec{x}\tan{x} \, dx =\sec{x} +C
  • \int \frac{1}{x}\, dx =\ln{|x|} +C
  • \int \frac{1}{x^2+1} \, dx =\tan^{-1}{x} +C
  • \int \frac{1}{\sqrt{1-x^2}} \, dx =\sin^{-1}{x} +C

Remark

The latter two, \int \frac{1}{x^2+1} \, dx =\tan^{-1}{x} +C and \int \frac{1}{\sqrt{1-x^2}} \, dx =\sin^{-1}{x} +C, will be discussed in the later section, Trigonometric Substitution.

The veracity of these antiderivatives can be confirmed by taking the derivative. Let us explore a few examples and then discover a few rules of integration.

Example. Verify \int \sec{x}\tan{x} \, dx =\sec{x} +C.

    \begin{align*} \frac{d}{dx}\sec{x} &= \frac{d}{dx}\left(\cos{x}\right)^{-1}\\ &= -\left(\cos{x}\right)^{-2}\left(\cos'{x}\right)\\ &= -\left(\cos{x}\right)^{-2}\left(-\sin{x}\right)\\ &= \sec{x}\tan{x} \end{align*}

Here is another example.

Example. Verify \int \frac{1}{x}\, dx = \ln{|x|} +C.

    \begin{align*} \frac{d}{dx}\ln{|x|} &= \lim_{h \to 0} \frac{\ln|x+h|-\ln|x|}{h}\\ &= \lim_{h \to 0} \frac{\ln \left\lvert \frac{x+h}{x}\right\rvert}{h} \\ &= \lim_{h \to 0} \ln \left\lvert 1+\frac{h}{x}\right\rvert^{\frac{1}{h}}\\ &= \ln \lim_{h \to 0} \left\lvert 1+\frac{h}{x}\right\rvert^{\frac{x}{h}\frac{1}{x}}\\ &= \ln |e|^{\frac{1}{x}}\\ &= \frac{1}{x} \end{align*}

Recall the chain rule in the Derivatives section. The inverse of the chain rule in integration is called the substitution rule. That is,

Theorem. (Substitution Rule)
If u=g(x) is a differentiable function with a range, an interval I, and f is continuous on I, then
    \[\int f(g(x))g'(x)\, dx=\int f(u)\,du\]

Let us illustrate with an example.

Example. Find \int 2x\cos{x^2}\,dx.

Let u=x^2. Then, du=2x\,dx. Then,

    \begin{align*} \int 2x\cos{x^2}\,dx &= \int \cos{u}\,du\\ &= \sin{u}+C\\ &= \sin{x^2}+C \end{align*}

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge connecting differential and integral calculus, and is divided into 2 parts. The first part establishes the interconnection between the two, that one of the antiderivates, denoted by F, of an integrable function f can be obtained as a definite integral of f. The second part, called Newton-Leibniz axiom, a definite integral of f can be computed by any one antiderivative F of its infinitely many antiderivatives.

Theorem. (Fundamental Theorem of Calculus)
Part 1.
If f is continuous on [a,b], then the function F defined by
    \[ F(x)=\int_{a}^{x}f(t)\,dt\]
is continuous on [a,b] and differentiable on (a,b), and
    \[F'(x)=f(x)\]

Part 2.
If f is continuous on [a,b], then
    \[ \int_{a}^{b}f(x)\,dx =F(b)-F(a)\]
where F'=f.

Integration by Parts

Recall the product rule of integration. That is,

    \begin{align*} \frac{d}{dx}\left[f(x)g(x)\right] &=f(x)\frac{d}{dx}g(x)+g(x)\frac{d}{dx}f(x)\\ &=f(x)g'(x)+f'(x)g(x) \end{align*}

Then, by integrating both sides,

    \begin{align*} f(x)g(x) &= \int \left[f(x)g'(x)+f'(x)g(x)\right]\,dx\\ &= \int f(x)g'(x)\,dx + \int f'(x)g(x)\,dx \end{align*}

Therefore,

    \[ \int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx\]

This can alternatively be expressed as follows, by letting u=f(x) and v=g(x):

    \[ \int u\,dv = uv - \int v\,du\]

Let us illustrate with an example.

Example. Find \int 2x\ln{x}\,dx.

Let u=\ln{x} and dv=2x \,dx.
Then, du=\frac{1}{x} and v=x^2.
Then,

    \begin{align*} \int 2x\ln{x}\,dx &= \int u\,dv\\ &= uv - \int v\,du\\ &= x^2\ln{x}-\int x\,dx\\ &= x^2\ln{x}-\frac{1}{2}x^2+C \end{align*}

Trigonometric Substitution

We can in fact inversely substitute x with another function g(t), so that

    \[ \int f(x)\,dx=\int f \left(g(t)\right)g'(t)\,dt\]

This line of trick or technique is useful whenever we encounter the following cases:

  • \sqrt{a^2-x^2}. \quad Let x=a \sin \theta for \theta \in \left[ -\frac{\pi}{2},\frac{\pi}{2} \right]
  • \sqrt{a^2+x^2}. \quad Let x=a \tan \theta for \theta \in \left( -\frac{\pi}{2},\frac{\pi}{2} \right)
  • \sqrt{x^2-a^2}. \quad Let x=a \sec \theta for \theta \in \left[ 0,\frac{\pi}{2} \right) or \theta \in \left[ \pi,\frac{3\pi}{2} \right)

It would be easier to illustrate with examples.

Example. Evaluate \int \frac{dx}{x^2 \sqrt{1+x^2}}.

Let x=\tan\theta for \theta \in \left( -\frac{\pi}{2},\frac{\pi}{2} \right). Then, dx=\sec^2{\theta}\,d\theta. Then,

    \begin{align*} \int \frac{dx}{x^2 \sqrt{1+x^2}} &= \int \frac{\sec^2{\theta}}{\tan^2 \theta \sqrt{1+\tan^2 \theta}} d\theta\\ &= \int \frac{\sec^2{\theta}}{\tan^2 \theta \sec{\theta}} d\theta\\ &= \int \frac{\cos{\theta}}{\sin^2{\theta}} d\theta \end{align*}

Now, let \sin{\theta}=t. Then, \cos{\theta}\,d\theta=dt. Then,

    \begin{align*} \int \frac{\cos{\theta}}{\sin^2{\theta}} d\theta &= \int \frac{dt}{t^2}\\ &= -\frac{1}{t}+C\\ &= -\frac{1}{\sin{\theta}}+C\\ &= -\frac{\sqrt{1+x^2}}{x}+C \end{align*}

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