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A Brief Introduction to Calculus: Part 2.1 - Definite and Indefinite Integrals

  • Math
  • Calculus
  • Further Math
Published on Thu Aug 04 2022. Views
An introduction to basic concepts in Calculus; mostly based on the IGCSE Further Math textbook.

Introduction

This series would primarily focus on the three fundamental areas of calculus - derivatives, limits, and integration, looking briefly at some of the different concepts to get started with calculus.; thus no prior knowledge of calculus is needed.

This is the first chapter of the second part of the series, and it is about definite and indefinite integrals.

Further content on this topic will be published in separate articles for performance reasons.

So now, let's sit back, relax, and have some fun!

Quick Access

Integration

The process of integration is essentially the reverse operation of differentiation. When differentiating a function, its polynomial degree is decreased to, for instance, work out the rate of change; whereas integration reverses this process and increments the polynomial degree.

An integral is also the anti-derivative of a function.

There are two types of integrals:

  • Indefinite integral
  • Definite integral

Indefinite Integrals

An indefinite integral is written in the form of ∫y dx\int y\,dx, where yy is the expression to be integrated (a.k.a the integrand) and xx is the variable to integrate with respect to.

For example, ∫2x dx\int 2x\, dx, is an indefinite integral whose integrand is 2x2x and the variable to integrate is xx.

Since an integral produces another function, by convention, that new function, i.e. the anti-derivative is represented by F(x)F(x). In other words:

∫f(x) dx=F(x)+C\begin{align*} \int f(x)\, dx = F(x) + C \end{align*}

where CC is some constant

As said earlier, integration is the reverse process of differentiation, so sometimes it will attempt to construct the function that's being differentiated.

For instance, if f(x)=x2f(x) = x^2, then the derivative of f(x)f(x) is given by f′(x)=2xf'(x) = 2x. However, let's say that we have another function g(x)=x2+5g(x) = x^2 + 5, its derivative would also be g′(x)=2xg'(x) = 2x. And hence with these two functions as integrands, we would obtain the same integral:

∫2x dx\begin{align*} \int 2x\, dx \end{align*}

Thus, if you were asked to find f(x)f(x) given that f′(x)=2xf'(x) = 2x, you cannot give a definite and single answer, because as seen above, f′(x)=g′(x)f'(x) = g'(x) but f(x)≠g(x)f(x) \not= g(x); and this is due to the constant in g(x)g(x), which is eliminated in the process of differentiation.

However, we can say that:

∫2x dx=x2+C\begin{align*} \int 2x\, dx = x^2 + C \end{align*}

Where CC is a possible constant ignored/removed when differentiating; it could be anything, but without further information, we won't be able to find CC.

Rules for Integration

Let's discuss some basic rules for integration before moving on to definite integrals.

Bear in mind that integration is the inverse operation, so you can verify them yourself if you want to see why these rules make sense.

Just like differentiating, you will need to integrate term by term.

Note that it's also very important to pay attention to the part of dxdx, because it indicates what variable you should integrate with respect to.

When applying the following rules to all the terms in a function, the constant CC is only added at the end and doesn't need to be included in every integrated term.

Powers of xx

∫xn dx=xn+1n+1+C   (n≠−1)\begin{align*} \int x^n\, dx = \frac{x^{n + 1}}{n + 1} + C \space\space\space (n \not= -1) \end{align*}

Examples

  1. Find ∫3x2 dx\int 3x^2\, dx
∫3x2 dx=33x3+C=x3+C\begin{align*} &\int 3x^2\, dx \\ =& \frac{3}{3}x^3 + C \\ =& x^3 + C \end{align*}
  1. Find ∫x4 dx\int x^4\, dx
∫x4 dx=x55+C\begin{align*} \int x^4\, dx = \frac{x^5}{5} + C \end{align*}
  1. Find ∫(x+2x3) dx\int \left(\sqrt{x} + 2x^3\right)\, dx
∫(x12+2x3) dx=23x32+12x4+C\begin{align*} & \int (x^{\frac{1}{2}} + 2x^3)\, dx \\ =& \frac{2}{3}x^\frac{3}{2} + \frac{1}{2}x^4 + C \end{align*}

A Single Constant

∫a dx=ax+C\begin{align*} \int a\, dx = ax + C \end{align*}

Example

∫3 dx=3x+C\begin{align*} \int 3\, dx = 3x + C \end{align*}

The Exponential Function

∫eax dx=1aeax+C\begin{align*} \int e^{ax}\, dx = \frac{1}{a}e^{ax} + C \end{align*}

Example

Find ∫5e−3x dx\int 5e^{-3x}\, dx

∫5e−3x dx= −15e−3x+C\begin{align*} & \int 5e^{-3x}\, dx \\ =&\space-15e^{-3x} + C \end{align*}

Sine & Cosine

∫sin⁡(ax) dx=−1acos⁡(ax)+C∫cos⁡(ax) dx=1asin⁡(ax)+C\begin{align*} &\int \sin(ax)\, dx = -\frac{1}{a}\cos(ax) + C \\ \\ &\int \cos(ax)\, dx = \frac{1}{a}\sin(ax) + C \\ \end{align*}

Example

find I=∫(2cos⁡(x)+3e−x−sin⁡(2x)) dxI = \int \left(2\cos(x) + 3e^{-x} - \sin(2x)\right)\, dx

∫2cos⁡(x) dx=2sin⁡(x)∫3e−x dx=−3e−x∫−sin⁡(2x) dx=−(−12cos⁡(2x))=12cos⁡(2x)\begin{align*} \int 2\cos(x)\, dx &= 2\sin(x) \\ \int 3e^{-x}\, dx &= -3e^{-x} \\ \int -\sin(2x)\, dx &= -(-\frac{1}{2}\cos(2x)) \\ &= \frac{1}{2}\cos(2x) \\ \\ \end{align*}
I=2sin⁡(x)−3e−x+12cos⁡(2x)+CI = 2\sin(x) - 3e^{-x} + \frac{1}{2}\cos(2x) + C

Inverse xx

For the powers of xx, there is one exception, and that is ax\frac{a}{x}, and xx needs to have a power of exactly −1-1 when it's rewritten as being multiplied by a. In other words, this doesn't include scenarios like ∫1x2 dx\int \frac{1}{x^2}\, dx

∫ax dx=aln⁡(∣x∣)+C\begin{align*} \int \frac{a}{x}\, dx = a\ln(|x|) + C \end{align*}

This is because:

∫1x dx=ln⁡(∣x∣)+C\begin{align*} \int \frac{1}{x}\, dx = \ln(|x|) + C \end{align*}

and:

∫ax dx=a⋅∫1x dx= aln⁡(∣x∣)+C\begin{align*} &\int \frac{a}{x}\, dx = a \cdot \int \frac{1}{x}\, dx \\ =\,& a\ln(|x|) + C \end{align*}

Example

Evaluate ∫4x dx\int \frac{4}{x}\, dx

∫4x dx=4ln⁡(x)+C\begin{align*} \int \frac{4}{x}\, dx = 4\ln(x) + C \end{align*}

Definite Integrals

Unlike indefinite integrals, these integrals have a limit of integration, composed of an upper and a lower bound. Moreover, they compute a definite value instead of producing a function.

A definite integral is written as follows:

∫aby dx\int_{a}^{b} y\, dx

What are aa and bb then?

bb represents the upper limit of integration, and aa represents the lower limit.

As discussed earlier, a function produced by an indefinite integral is written as F(x)F(x).

However, a definite integral is simply the value of F(b) - F(a)$.

Hence:

∫abf(x) dx=[ F(x) ]ab=F(b)−F(a)\begin{align*} & \int_{a}^{b} f(x) \, dx \\ =& \left[\,F(x)\,\right]_{a}^{b} \\ =& F(b) - F(a) \end{align*}

Notice that the constant CC is not included in the expressions, that's because when F(b)+C−(F(a)+C)F(b) + C - (F(a) + C), the constant cancels out.

Example: evaluate ∫14[x3+4x2−7x+13] dx\int_{1}^{4} \left[x^3 + 4x^2 - 7x + 13\right]\, dx

∫14[x3+6x2−7x+13] dx= [x44+2x3−72x2+13x]14= 188−1134= 17614\begin{align*} &\int_{1}^{4} \left[x^3 + 6x^2 - 7x + 13\right]\, dx \\ =&\, \left[\frac{x^4}{4} + 2x^3 - \frac{7}{2}x^2 + 13x\right]_{1}^{4} \\ =&\, 188 - 11\frac{3}{4} \\ =&\, 176\frac{1}{4} \end{align*}

Conclusion

This blog introduced a few basic concepts of definite and indefinite integrals, and more content will be covered in future blogs of this part.

Thanks for reading!