An Introduction to Calculus – Integration and Applications – A Guide to Mathematical Studies (Archive)

Continuing the topic of calculus, we now introduce another function – integration.

Imagine we want to find the area under the curve of the equation $y=3x^2-4x+6$ in the range $0\le{x}\le2$ . As we know, the area cannot be calculated through normal means, as indicated by the graph below.

In order to find the area of the curve (shown by the highlighted green area), we can divide the area into a sum of trapeziums, with side lengths of $\Delta{x}$ for the $x$ axis. As $\Delta{x}$ tends to 0, we obtain a closer and closer approximation to the actual area under the curve. This can be generalised with the following equation :

$\int_a^bf(x)dx=\lim_{\Delta{x}\rightarrow0}\Delta{x}\sum^{\infty}_{i=0}f(x_i)$

In this case, $a \text{ and } b$ are the lower bounds and upper bounds of the integral respectively, and $\int f(x)\text{ }dx$ is the proper notation of an integral, with $\int$ being the integral operator and $dx$ indicating the variable to integrate. This is the definition of the Riemann sum, which reduces the integral of a curve into a sum of areas. This also has a surprising connection with differentiation, where the integration operator is the inverse function of differentiation. This is known as the first fundamental theorem of calculus.

$\frac{d}{dx}\int{f(x)\text{ }dx=f(x)}$

In fact, we can easily prove this with the first principle of derivatives (see An Introduction to Calculus – Limits and Differentiation) given the second fundemental theorem of calculus, which is much more intuitive and hence its proof is left for readers.

Let $f(x)$ be the function which’s area we want to evaluate and $F(x)$ be the antiderivative (i.e. integral) of the function. The second fundemental theorem states the following :

$\int^b_a{f(x)}dx=F(b)-F(a)$

However, the function must be continuous on all intervals for the theorem to hold true.

Now we can prove the first fundemental theorem of calculus. By the above definition of Riemann sum, the area under the curve for one strip would be $\lim_{\Delta{x}\rightarrow0}\Delta{x}f(x)$ .

$\int^b_a{f(x)}dx=F(b)-F(a)$

$F(b)-F(a) = \lim_{\Delta{x}\rightarrow0}\Delta{x}f(x)$

$\lim_{\Delta{x}\rightarrow0}\frac{F(b)-F(a)}{\Delta{x}} = f(x)$

This is the exact definition of the derivative – hence by this, we have proven that in fact the integral of a function is an inverse function of differentiation. This means, in order to find the area of a certain function, we need to carry out an inverse operation of differentiation for the variable specified.

To demonstrate this, we will find the area of the above equation $y=3x^2-4x+6$ in the range $0\le{x}\le2$ .

Firstly, let us write this out in mathematical notation : $\int^2_0 3x^2-4x+6$ .
We can then use the power rule : $[x^3+2x^2+3x]_0^2$ , and by using the second fundamental theorem of calculus we get : $(2^3+2(2)^2+3(2))-(0^3-2(0)^2+3(0))$ which equals 22. Using integration, we have demonstrated how to find the area under the curve for elementary functions (you will learn how to find areas of integrals of non-elementary functions such as $\int_0^{\infty}\frac{\sin(x)}{x}$ with contours or Laplace transformation in Calc III).

However, for general cases where you only want to find the antiderivative, the method is the same except that a constant is added to the equation. This is explained by taking the derivative of the constant, which we know is 0. This constant compensates for the fact that it is lost during the action of taking a derivative. This is needed when solving for initial-value problems :

Example : Given $\frac{dy}{dx}=2xy$ and $y(0)=3$ , find $y$ in terms of $x$ .

Answer : We multiply both sides by $dx$ to get $dy = 2xy\, dx$ . We further separate variables to obtain $\frac{1}{y}\, dy = 2x\, dx$ .
Integrate both sides : $\int \frac{1}{y}\, dy= \int 2x\, dx$ , which gives us $\ln(y)=x^2+c$ (we combine the two constants from the integral into one), then solve for $y$ by raising both sides by $e$ , which gives us $e^{x^2+c}$ . We now substitute $x=0 \text{ \& } y=3$ : $3= e^{0^2+c} \Longrightarrow{e^c=3}\Longrightarrow{c=\ln(3)}$ .
Hence the solution is $y=e^{x^2+\ln(3)}$ , or in other terms $y=3e^{x^2}$ .

We can now apply this to solve real-life problems :

Edexcel AS Further Mechanics 2 – June 2022

For part (a), as we know $\frac{dv}{dt}=a$ , we take the differential of the velocity, namely $a = \frac12{6e^{2t}}$ . From this, it is trivial that $a =2v+1$ .
For part (b), simply substitute $t=0$ into the acceleration we calculated. This is easily shown to be 3.
For part (c), we know that $\frac{ds}{dt}=v$ , and thus $\int{v}\, dt=s$ . However, this is a definite integral, but we don’t have the time as the upper or lower bound. What could we do?

Fortunately, we have the initial speed and final speed provided. By plugging these into our equation for velocity, we can find the respective time.

$1 = \frac12({3e^{2t}-1}) \Longrightarrow {t=0}$

$4 = \frac12{({3e^{2t}-1})} \Longrightarrow {t=\frac{ln(3)}2}$

We now have the lower and upper bounds for our integral. Thus, we can integrate the equation :

$\int_0^{\frac{ln(3)}2}{\frac12({3e^{2t}-1}})\, dt$

$\frac12[\int_0^{\frac{ln(3)}2}{e}^{2t}\, dt-\int_0^{\frac{ln(3)}2}1\, dt]$

$\frac12[{\frac12}e^{2t}]_0^{\frac{ln(3)}2}-\frac12[t]_0^{\frac{ln(3)}2}$

Plugging in the values, we get $\frac32-\frac{1}{4}ln(3)$ .

We also need to understand a method called $u$ -subsitution for integration. We identify a proper parameter for $u$ , which we then subsitute into the equation to simply the expression. We also have to change the boundaries of integration if given. Below we shall demonstrate an example of this :

Example : Find $\int_1^e\frac{\ln(x)}{x}\, dx$

Answer : We notice that $\frac{\ln(x)}{x}$ can be written as $\ln(x)\cdot\frac1x$ , where $\frac{d}{dx} \ln(x)$ is $\frac1x$ . We carry out $u$ -subsitution for $u=\ln(x)$ , which gives us $du = \frac1x$ . We proceed to subsitute, remembering to change the boundaries. This is done by subsituting the $x$ value into the equation : $u=\ln(1)=0$ and $u=\ln(e)=1$ . Now we can integrate : $\int_0^{1}u\, du = [\frac12{u^2}]^{1}_0=\frac12$ . This can be applied in many scenarios, the most notable being trignometric subsitution.¹

Another rule that we must introduce is integration by parts, a very useful method for integrating products of two or more functions. It is stated as follows :

For two functions $u=u(x)$ and $g=g(x)$ ,

$\int{u\, dv} = uv-\int{v\, du}$

This is proven easily using derivatives – we first consider the product rule, which states $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$ for $y=uv$ . We rearrange to get $u\frac{dv}{dx} = \frac{dy}{dx} - v\frac{du}{dx}$ , which we integrate with respect to $x$ to obtain $\int{u\frac{dv}{dx}}{dx}=y-\int{v\frac{du}{dx}}{dx}$ . We subsitute our original equation for $y$ and simplify to finally obtain $\int{u\, dv} = uv-\int{v\, du}$ . Let us see another example for this :

Example : Integrate $\int{x}e^x\, dx$ .

Answer : We integrate this using integration by parts, namely we let $x$ be $u$ and $e^x$ be $\, dv$ (can you think of a reason why this order but not the other way round?), which by applying the formula becomes $xe^x-\int{e^x{\frac{d}{dx}x}}=xe^x-\int{e^x\, dx}=xe^x-e^x+C$ .

In conclusion, we’ve explored the fascinating world of calculus, specifically delving into the realm of integration and its practical applications. We began by grasping the fundamental concept of integration as a tool for finding areas, accumulation, and continuous change. Through the lenses of definite and indefinite integrals, we’ve harnessed this mathematical technique to solve real-world problems ranging from physics and engineering to economics and biology.

Calculus, particularly integration, is more than a mathematical exercise; it is a versatile and indispensable framework for understanding the dynamics of our complex world. From determining the work done in physics to estimating the total profit in economics, integration serves as a bridge between abstract mathematical concepts and their tangible real-world consequences. I wish to see you in our next installment, and if you liked this, consider subscribing to my newsletter!

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An Introduction to Calculus – Integration and Applications

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