Differential and Integral Calculus

For example, \(f\colon \mathbb{R}\to\mathbb{R}\), \(f(x)=x^2\), has codomain \(\mathbb{R}\), but its image is \(f[\mathbb{R} ] =[0,\infty[\).

The function in the previous example can also be defined as \(f\colon \mathbb{R}\to [0,\infty[\), \(f(x)=x^2\), and then the codomain is the same as the image. In principle, this modification can always be done, but it is not reasonable in practice.

Example: Try to do the same for \(f\colon \mathbb{R}\to\mathbb{R}\), \(f(x)=x^6+x^2+x\), \(x\in\mathbb{R}\).

• If the domain \(A\subset \mathbb{R}\) then \(f\) is a function of one (real) variable: the main object of study in this course.

• If \(A\subset \mathbb{R}^n\), \(n\ge 2\), then \(f\) is a function of several variables (a multivariable function)

Observe: A function becomes surjective if all redundant points of the codomain are left out. A function becomes injective if the domain is reduced so that no value of the function is obtained more than once.

Another way of defining these concepts is based on the number of solutions to an equation:

The inverse satisfies \(f^{-1}(f(a))=a\) for all \(a\in A\) and \(f(f^{-1}(b))=b\) for all \(b\in B\).

The graph of the inverse is the mirror image of the graph of \(f\) with respect to the line \(y=x\): A point \((a,b)\) lies on the graph of \(f\) \(\Leftrightarrow\) \(b=f(a)\) \(\Leftrightarrow\) \(a=f^{-1}(b)\) \(\Leftrightarrow\) the point \((b,a)\) lies on the graph of \(f^{-1}\). The geometric interpretation of \((a,b)\mapsto (b,a)\) is precisely the reflection with respect to \(y=x\).

If \(A \subset \mathbb{R}\) and \(f\colon A\to \mathbb{R}\) is strictly monotone, then the function \(f\colon A \to f[A]\) has an inverse.

If here \(A\) is an interval and \(f\) is continuous, then also \(f^{-1}\) is is continuous in the set \(f[A]\).

Proof.

Note. \(f'(f^{-1}(x))\) is the derivative of \(f\) at the point \(f^{-1}(x)\).

Trigonometric functions

• Unit of measurement of an angle = rad: the arclength of the arc on the unit circle, that corresponds to the angle.

• \(\pi\) rad = \(180\) degrees, i.e. \(1\) rad = \(180/\pi \approx 57,\! 3\) degrees

• The functions \(\sin x, \cos x\) are defined in terms of the unit circle so that \((\cos x,\sin x)\), \(x\in [0,2\pi]\), is the point on the unit circle corresponding to the angle \(x\in\mathbb{R}\), measured counterclockwise from the point \((1,0)\). \[\tan x = \frac{\sin x}{\cos x}\ (x\neq \pi /2 +n\pi),\] \[\cot x = \frac{\cos x}{\sin x}\ (x\neq n\pi)\]

• Periodicity: \[\sin (x+2\pi) = \sin x,\ \cos (x+2\pi)=\cos x,\] \[\tan (x+\pi) = \tan x\]

Basic properties (from the unit circle!)

• \(\sin 0 = 0\), \(\sin (\pi/2)=1\)

• \(\cos 0=1\), \(\cos (\pi/2)= 0\)

• Parity: \(\sin\) and \(\tan\) are odd functions, \(\cos\) is an even function: \[\sin (-x) = -\sin x,\] \[\cos(-x) = \cos x,\] \[\tan (-x) = -\tan x.\]

• \(\sin^2 x + \cos^2 x = 1\) for all \(x\in\mathbb{R}\)

  Proof: Pythagorean Theorem.

• Addition formulas:

  \(\sin (x+y) = \sin x \cos y +\cos x \sin y\)

  \(\cos (x+y) = \cos x \cos y -\sin x \sin y\)

Proof: Geometrically, or more easily with vectors and matrices.

Derivatives: \[ D(\sin x) = \cos x,\ \ D(\cos x) = -\sin x \]

Example

It follows that the functions \(y(t)=\sin (\omega t)\) and \(y(t)=\cos (\omega t)\) satisfy the differential equation \[ y''(t)+\omega^2y(t)=0, \] that models harmonic oscillation. Here \(t\) is the time variable and the constant \(\omega>0\) is the angular frequency of the oscillation. We will see later that all the solutions of this differential equation are of the form \[ y(t)=A\cos (\omega t) +B\sin (\omega t), \] with \(A,B\) constants. They will be uniquely determined if we know the initial location \(y(0)\) and the initial velocity \(y'(0)\). All solutions are periodic and their period is \(T=2\pi/\omega\).

Arcus functions

The trigonometric functions have inverses if their domain and codomains are chosen in a suitable way.

• The Sine function \[ \sin \colon [-\pi/2,\pi/2]\to [-1,1] \] is strictly increasing and bijective.

• The Cosine function \[ \cos \colon [0,\pi] \to [-1,1] \] is strictly decreasing and bijective.

• The tangent function \[ \tan \colon ]-\pi/2,\pi/2[\, \to \mathbb{R} \] is strictly increasing and bijective.

We consider areas of plane sets bounded by closed curves. In the more general cases, the concept of area becomes theoretically very difficult.

The area of a planar set is defined by reducing to the areas of simpler sets. The area cannot be "calculated", unless we first have a definition of "area" (although this is common practice in school mathematics).

A (simple) polygon is a plane set bounded by a closed curve that consists of a finite number of line segments without self-intersections.

For a plane set \(\color{red} D\) bounded by a closed curve we can construct inner polygons \(\color{blue}P_i\) and outer polygons \(P_o\): \(\color{blue}P_i\color{black} \subset \color{red}D\color{black}\subset P_o\).

A surprise: The condition that \(D\) is bounded by a closed curve (without self-intersections) does not guarantee that it has an area! Reason: The boundary curve can be so "wiggly", that it has positive "area". The first such example was constucted by [W.F. Osgood, 1903]:

Example

Derive the formula \(A=\pi R^2\) for a circle with radius \(R\) by choosing regular inscrided and circumscribed \(n\)-gons as inner and outer polygons, and let \(n\to\infty\).

The solution is a voluntary exercise, where you need the limit \[\lim_{x\to 0}\frac{\sin x}{x} = 1.\] Hint: Show that the inscribed and circumscribed areas are \[ \pi R^2\frac{\sin (2\pi/n)}{2\pi/n} \ \text{ and }\ \pi R^2\frac{\tan \pi/n}{\pi/n}.\]

Geometric interpretation: Let \(f\colon[a,b]\to\mathbb{R}\) be such that \(f(x)\ge 0\) for all \(x\in[a,b]\). How can we find the area of the region bounded by the function graph \(y=f(x)\), the x-axis and the two lines \(x=a\) and \(x=b\)?

The answer to this question is given by the definite integral \[\int_{a}^{b}f(x)\,dx\] Remark. The general definition of the integral does not necessitate the condition \(f(x)\ge 0\).

Integration of continuous functions

If \(f\) is a positive function then the upper sum represents the total area of the rectangles circumscribing the function graph and similarly the lower sum is the total area of the inscribed rectangles.

Remark. This definition of the integral is sometimes referred to as the Darboux integral.

For non-negative functions \(f\) this definition of the integral coincides with the idea of making the difference between the the areas of the circumscribed and the inscribed rectangles arbitrarily small by using ever finer partitions.

Proof.

Remark. This definition of the integral turns out to be equivalent to that of the Darboux integral i.e. a function is Riemann-integrable if and only if it is Darboux-integrable and the values of the two integrals are always equal.

Remark. Any interval \([a,b]\) can be partitioned into equally spaced subintervals by setting \(\Delta x = (b-a)/n\) and \(x_{k} = a + k\Delta x\).

Piecewise-defined functions

Example

Consider the function \(f\colon\left[-1,1\right]\) defined as \[ f(x) = \begin{cases} -1 &\text{ for }-1\le x<0 \\ 1 &\text{ for }0\le x\le 1. \end{cases} \] We can now integrate \(f\) as follows: \[ \int_{-1}^{1}f(x)\,dx = \int_{-1}^{0}f(x)\,dx + \int_{0}^{1}f(x)\,dx \] \[ =\int_{-1}^{0}(-1)\,dx + \int_{0}^{1}1\,dx = -1\cdot(-1-0) + 1\cdot(1-0) = 2. \]

Important properties

Fundamental theorem of calculus

Proof.

Proof.

If \(F'(x)=f(x)\) on some open interval then \(F\) is the antiderivative (or the primitive function) of \(f\). The fundamental theorem of calculus guarantees that for every continuous function \(f\) there exists an antiderivative \[ F(x) = \int_{a}^{x}f(t)\,dt. \] The antiderivative is not necessarily expressible as a combination of elementary functions even if \(f\) were an elementary function, e.g. \(f(x) = e^{-x^{2}}\). Such primitives are called nonelementary antiderivatives.

Proof.

Proof.

Proof.

Constant Functions

Given the constant function \(f(x) = c,\,c\in\mathbb{R}\). The integral \(\int\limits_a^b f(x)\,\mathrm{d} x = \int\limits_a^b c \, \mathrm{d}x\) has to be determined now.

Solution by finding a antiderivative

From the previous chapter it is known that \(g(x) = c\cdot x\) gives \(g'(x) = c\). This means that \(c \cdot x\) is an antiderivative for \(c\). So the following applies \[\int\limits_a^b c \, \mathrm{d}x = [c \cdot x]_{x=a}^{x=b} = c\cdot b - c \cdot a = c \cdot (b-a).\]

Remark: Of course, a function \(h(x) = c \cdot x + d\) would also be an antiderivative of \(f\), since the constant \(d\) is omitted in the derivation. For sake of simplicity \(c \cdot x\) can be used, since \(d\) can be chosen as \(d=0\) for definite integrals.

Solution by geometry

The area under the constant function forms a rectangle with height \(c\) and length \(b-a\). Thus the area is \(c \cdot (b-a)\) and this corresponds to the solution of the integral. Illustrate this remark by a sketch.

Linear functions

Given is the linear function \( f(x) = mx\). We are looking for the integral \(\int\limits_a^b f(x)\, \mathrm dx=\int\limits_a^b mx\, \mathrm dx\).

Solve by finding a antiderivative

The antiderivative of a linear function is in any case a quadratic function, since \(\frac{\mathrm d x^2}{\mathrm dx} = 2x \). The derivative of a quadratic function results in a linear function. Here, it is important to consider the leading factor as in \[\frac{\mathrm d (m \cdot \frac{1}{2} \cdot x^2)}{\mathrm dx} = mx.\] Thus the result is \[\int\limits_a^b mx \mathrm{d}x = \left[\frac{m}{2}x^2 \right]_{x=a}^{x=b}= \frac{m}{2}b^2 - \frac{m}{2}a^2. \]

Solving by geometry

The integral \(\int\limits_a^b mx\, \mathrm dx\) can be seen geometrically, as subtracting the triangle with the edges \((0|0)\), \((a|0)\) and \((a| ma)\) from the triangle with the edges \((0|0)\), \((b|0)\) and \((b| mb)\). Since the area of a triangle ist given by \(\frac{1}{2} \cdot \mbox{baseline} \cdot \mbox{height}\), the area of the first triangle \(\frac{1}{2}\cdot b \cdot mb = \frac{1}{2}mb^2\) and that of the second triangle is analogous \(\frac{1}{2}ma^2\). For the integral the result is \(\frac{m}{2}b^2 - \frac{m}{2}a^2\). This is consistent with the integral calculated using the antiderivative. Illustrate this remark by a sketch.

Power functions

In constant and linear functions we have already seen that the exponent of a function decreases by one when it is derived. So it has to get bigger when integrating. The following applies: \[\frac{\mathrm d x^n}{\mathrm dx} = n \cdot x^{n-1}. \] It follows that the antiderivative for \(x^n\) must have the exponent \(n+1\), \[\frac{\mathrm d x^{n+1}}{\mathrm dx} = (n+1) \cdot x^n.\] By multiplying the last equation with \(\frac{1}{n+1}\) we get \[\frac{\mathrm d}{\mathrm dx}\frac{1}{n+1} x^{n+1} = \frac{n+1}{n+1} \cdot x^n = x^n.\] Finally the antiderivative is \(\int x^n\, \mathrm dx = \frac{1}{n+1} x ^{n+1}+c,\,c\in\mathbb{R}\).

Examples

• \(\int x^2\, \mathrm dx = \frac{1}{3} x^3 +c,\,c\in\mathbb{R}\)
• \(\int x^3\, \mathrm dx = \frac{1}{4} x^4 +c,\,c\in\mathbb{R}\)
• \(\int x^{20}\, \mathrm dx = \frac{1}{21} x^{21} +c,\,c\in\mathbb{R}\)

The formula \(\int x^n\, \mathrm dx = [\frac{1}{n+1} x^{n+1}] \) is also valid, if the exponent of the function is a real number and not equal \(-1\).

Examples

• \(\int x^{2,7}\, \mathrm dx = \frac{1}{3,7} x^{3,7}+c,\,c\in\mathbb{R}\)
• \(\int \sqrt{x}\, \mathrm dx = \int x^\frac{1}{2} = \frac{2}{3} x^{\frac{3}{2}}+c,\,c\in\mathbb{R}\)
• But: For \(x\gt0\) applies\(\int x^{-1}\,\mathrm d x=\ln(x)+c,\,c\in\mathbb{R}.\)

Natural Exponential function

The natural exponential function \(f(x) = e^x\) is one of the easiest function to differentiate and integrate. Since the derivation of \(e^x\) results in \(e^x\), it follows \[\int e^x\, \mathrm dx = e^x +c, \, c\in \mathbb{R}.\]

Example 1

Determine the value of the integral \(\int_0^1 e^z \,\mathrm{d} z\).

\[\int\limits_0^1e^z\,\mathrm{d}z= e^z\big|_{z=0}^{z=1}=e^1-e^0=e-1.\]

Example 2

Determine the value of the integral \(\int_0^b e^{\alpha t} \,\mathrm{d} t\). Using the same considerations as above we get \[\int\limits_0^b e^{\alpha t}\,\mathrm{d}t= \frac{1}{\alpha}e^{\alpha t}\big|_{t=0}^{t=b} =\frac{1}{\alpha}\left(e^{\alpha b}-e^0\right)=\frac{1}{\alpha}\left(e^{\alpha b}-1\right).\] Important is here, that we have to use the factor \(\frac1{\alpha}\).

Natural Logarithm

The derivative of the natural logarithmic function is \(\ln'(x) =\frac{1}{x}\) for \(x\gt0\). It even applies \(\ln'(x) =\frac{1}{x}\) to \(x<0\). These results together result in for the antiderivative of \(\frac{1}{x}\)

\[\int \frac{1}{x}\,\mathrm{d}x = \ln\left(|x|\right) +c , c\in\mathbb{R}.\]

An antiderivative can be specified for the natural logarithm: \[\int \ln(x)\,\mathrm{d}x = x\ln(x) - x + c ,\, c\in\mathbb{R}.\]

Trigonometric function

The antiderivatives of \(\sin(x)\) and \(\cos(x)\) also result logically if you derive "backwards". We have \[\int \sin(x)\, \mathrm dx = -\cos(x)+c,\,c\in\mathbb{R},\] since \( (-\cos(x))' =-(-\sin(x))=\sin(x).\) Furthermore we know \[\int \cos(x)\, \mathrm dx = \sin(x)+c,\,c\in\mathbb{R},\] since \((\sin(x))' = \cos(x) \) applies.

Example 1

Which area is covered by the sine on the interval \([0,\pi]\) and the \(x\)-axis? To determination the area we simply have to evaluate the integral \[\int_0^{\pi} \sin(\tau) \, \mathrm{\tau}.\] That means \[\int_0^{\pi} \sin(\tau) \, \mathrm{d}\tau = \left[-\cos(\tau)\right]_{\tau=0}^{\tau=\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2.\] Again make a sketch for this example.

Example 2

How can the integral \(\int \cos(\omega t +\phi)\,\mathrm{d}t\) be expressed analytically?

To determine the integral we use the antiderivative of the cosine: \(\sin'(x) = \cos(x)\). However, the inner derivativ has to be considered in the given function and thus we get \[\int \ \cos(\omega t +\phi)\,\mathrm{d}t =\frac{1}{\omega}\sin(\omega t+\phi)+c,\,c\in\mathbb{R}.\]

Example 1

Evaluate the integrals \(\displaystyle \int_{-1}^{1}e^{-x}\,dx\) and \(\displaystyle\int_{0}^{1}\sin(\pi x)\,dx\).

Solution. The antiderivative of \(e^{-x}\) is \(-e^{-x}\) so we have that \[\int_{-1}^{1}e^{-x}\,dx = -e^{-1}+e^{1} = 2\sinh1.\] The antiderivative of \(\sin(\pi x)\) is \(-\frac{1}{\pi}\cos(\pi x)\) and thus \[\int_{0}^{1}\sin(\pi x)\,dx = -\frac{1}{\pi}(\cos\pi-\cos0) = \frac{2}{\pi}.\]

Example 2

Evaluate the integral \(\displaystyle\int_{0}^{1}\frac{x}{\sqrt{25-9x^{2}}}\,dx\).

Solution. The antiderivative might look something like \(F(x)=a(25-9x^{2})^{1/2}\), where we can find the factor \(a\) through differentiation: \[D\big(a(25-9^{2})^{1/2}\big) = a\cdot\frac{1}{2}\cdot(-18x)(25-9x^{2})^{-1/2} = \frac{-9ax}{\sqrt{25-9x^{2}}}\] hence if \(a=-1/9\) we get the correct antiderivative. Thus \[\int_{0}^{1}\frac{x}{\sqrt{25-9x^{2}}}\,dx = -\frac{1}{9}\cdot(25-9x^{2})^{1/2}\Big|_{x=0}^{x=1} = -\frac{1}{9}(\sqrt{16}-\sqrt{25}) = \frac{1}{9}.\] This integral can also be solved using integration by substitution; more on this method later.

Suppose that \(f\) and \(g\) are piecewise continuous functions. The area of a region bounded by the graphs \(y=f(x)\), \(y=g(x)\) and the vertical lines \(x=a\) and \(x=b\) is given by the integral \[A=\int_{a}^{b}|f(x)-g(x)|\,dx.\]

Especially if \(f\) is a non-negative function on the interval \([a,b]\) and \(g(x)=0\) for all \(x\) then the integral \[A=\int_{a}^{b}f(x)\,dx\] is the area of the region bounded by the graph \(y=f(x)\), the \(x\)-axis and the vertical lines \(x=a\) and \(x=b\).

The arc length \(\ell\) of a planar curve \(y=f(x)\) between points \(x=a\) and \(x=b\) is given by the integral \[\ell = \int_{a}^{b}\sqrt{1+f'(x)^{2}}\,dx.\]

Heuristic reasoning: On a small interval \(\left[x,x+\Delta x\right]\) the arc length of the curve between \(y=f(x)\) and \(y=f(x+\Delta x)\) is approximately \[\Delta s \approx \sqrt{\Delta x^{2} + \Delta y^{2}} = \Delta x\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^{2}} \approx \Delta x\sqrt{1+f'(x)^{2}}.\]

The area of a surface generated by rotating the graph \(y=f(x)\) around the \(x\)-axis on the interval \(\left[a,b\right]\) is given by \[A = 2\pi\int_{a}^{b}|f(x)|\sqrt{1+f'(x)^{2}}\,dx.\] Heuristic reasoning: An area element of the surface is approximately \[\Delta A \approx \text{perimeter}\cdot\text{length} = 2\pi|f(x)|\cdot\Delta s.\]

Solid of revolution

Suppose that the cross-sectional area of a solid is given by the function \(A(x)\) when \(x\in\left[a,b\right]\). Then the volume of the solid is given by the integral \[V = \int_{a}^{b}A(x)\,dx.\] If the graph \(y=f(x)\) is rotated around the \(x\)-axis between the lines \(x=a\) and \(x=b\) the volume of the generated figure (the solid of revolution) is \[V = \pi\int_{a}^{b}f(x)^{2}\,dx.\] This follows from the fact that the cross-sectional area of the figure at \(x\) is a circle with radius \(f(x)\) i.e. \(A(x)=\pi f(x)^{2}\).

More generally: Let \(0\le g(x)\le f(x)\) and suppose that the region bounded by \(y=f(x)\) and \(y=g(x)\) and the lines \(x=a\) and \(x=b\) is rotated around the \(x\)-axis. The volume of this solid of revolution is \[V = \pi\int_{a}^{b}\big(f(x)^{2}-g(x)^{2}\big)\,dx.\]

One limitation of the improper integration is that the limit must be taken with respect to one endpoint at a time.

Example

\[\int_{0}^{\infty}\frac{dx}{\sqrt{x}(1+x)} = \int_{0}^{1}\frac{dx}{\sqrt{x}(1+x)} + \int_{1}^{\infty}\frac{dx}{\sqrt{x}(1+x)}\] Provided that both of the integrals on the right-hand side converge. If either of the two is divergent then so is the integral.

Example

Find the value of \(\displaystyle\int_{0}^{\infty}e^{-x}\,dx\).

Solution. Notice that \[\int_{0}^{R}e^{-x}\,dx = \left(-e^{-x}\right)\Bigg|_{x=0}^{R}=1-e^{-R}\to 1\] as \(R\to\infty\). Thus the improper integral converges and \[\int_{0}^{\infty}e^{-x}\,dx = 1.\]

However, this doesn't apply in general. For example, let \(f(x)=x\). Note that even though \[ \int_{-R}^{R}f(x)\,dx = \int_{-R}^{R}x\,dx = \frac{R^2}{2} - \frac{(-R)^2}{2} = 0 \] for all \(R\in\mathbb{R}\) the improper integral \[ \int_{-\infty}^{\infty}x\,dx = \lim_{R\to\infty}\int_{-R}^{0}x\,dx + \lim_{R\to\infty}\int_{0}^{R}x\,dx = \lim_{R\to\infty}-\frac{(-R)^2}{2} + \lim_{R\to\infty}\frac{R^2}{2} = \infty - \infty \] does not converge.

Example

Find the value of the improper integral \(\displaystyle\int_{0}^{1}\frac{dx}{\sqrt{x}}\).

Solution. We get \[\int_{\epsilon}^{1}\frac{dx}{\sqrt{x}} = \left(2\sqrt{x}\right)\Bigg|_{x=\epsilon}^{x=1} = 2-2\sqrt{\epsilon} \to 2,\] as \(\epsilon\to0+\). Thus the integral converges and its value is \(2\).

Integration techniques

Logarithmic integration

Given a quotient of differentiable functions, we know to apply the quotient rule. However, this is not so easy with integration. Here only for a few special cases we will state rules in this chapter.

Logarithmic integration As we already know the derivative of \(\ln(x)\), i.e. the natural logarithm to the base \(e\), equal to \(\frac{1}{x}\). According to the chain rule the derivative of differentiable function with positive function values is \(f\,:\,\frac{\mathrm d}{\mathrm dx} \ln (f(x)) = \frac{f'(x)}{f(x)}\). This means that for a quotient of functions where the numerator is the derivative of the denominator yields the rule: \begin{equation} \int \frac{f'(x)}{f(x)}\, \mathrm{d} x= \ln \left(|f(x)|\right) +c,\,c\in\mathbb{R}.\end{equation} Using the absolute value of the function is important, since the logarithm is defined on \(\mathbb{R}^+\).

Examples

• \(\int \frac{1}{x}\, \mathrm dx = \int \frac{x'}{x} \mathrm\, dx = \ln(|x|) +c,\,c\in\mathbb{R} \).
  
  
• \(\int \frac{3x^2 + 17}{x^3 +17x - 15}\, \mathrm dx = \ln(|x^3 + 17x - 15|)+c,\,c\in\mathbb{R}\).
  
  
• \(\int \frac{\cos(x)}{\sin(x)}\,\mathrm dx = \ln(|\sin(x)|)+c,\,c\in\mathbb{R}\).

Integration of rational functions - partial fraction decomposition

The logarithmic integration works well in special cases of broken rational functions where the counter is a multiple of the derivation of the denominator. However, other cases can sometimes be traced back to this. This method is called partial fractional decomposition, which represents rational functions as the sum of proper rational functions.

Example 1

The function \(\frac{1}{1-x^2}\) cannot be integrated at first glance. However, the denominator \(1-x^2\) can be written as \((1-x)(1+x)\) and the function can finally reads as \(\dfrac{1}{1-x^2} = \dfrac{\frac{1}{2}}{1+x} + \dfrac{\frac{1}{2}}{1-x}\) by partial fraction decomposition. This expression can be integrated, as demonstrated now: \begin{eqnarray} \int \dfrac{1}{1-x^2} \,\mathrm dx &= & \int \dfrac{\frac{1}{2}}{1+x} + \dfrac{\frac{1}{2}}{1-x}\, \mathrm dx \\ & =& \frac{1}{2} \int \dfrac{1}{1+x}\, \mathrm dx - \frac{1}{2} \int \dfrac{-1}{1-x}\, \mathrm dx\\ & = &\frac{1}{2} \ln|1+x| +c_1 - \frac{1}{2} \ln|1-x| +c_2\\ &= &\frac{1}{2} \ln \left|\dfrac{1+x}{1-x}\right|+c,\,c\in\mathbb{R}. \end{eqnarray} This procedure is now described in more detail for some special cases.

Case 1: \(Q(x)=(x-\lambda_1)(x-\lambda_2)\) with \(\lambda_1\ne\lambda_2\). In this case, \(R\) has the representation \(R(x) = \frac{ax+b}{(x-\lambda_1)(x-\lambda_2)}\) and can be transformed to \[\frac{ax+b}{(x-\lambda_1)(x-\lambda_2)} = \frac{A}{(x-\lambda_1)}+\frac{B}{(x-\lambda_2)}.\] By multiplying with \((x-\lambda_1)(x-\lambda_2)\) it yields ot \[ax+b = A(x-\lambda_1) + B(x-\lambda_2) = \underbrace{(A+B)}_{\stackrel{!}{=}a}x + \underbrace{(-A\lambda_1-B\lambda_2)}_{\stackrel{!}{=}b}.\]

\(A\) and \(B\) are now obtained by the method of equating the coefficients.

Example 2

Determe the partial fraction decomposition of \(\frac{2x+3}{(x-4)(x+5)}\).

Start with the equation \[\frac{2x+3}{(x-4)(x+5)} = \frac{A}{(x-4)}+\frac{B}{(x+5)}\] to get the parameters \(A\) and \(B\). Multiplication by \({(x-4)(x+5)}\) leads to \[2x+3 = A(x+5)+B(x-4) = (A+B)x +5A -4B.\] Now we get the system of linear equations

\begin{eqnarray}A+B & = & 2 \\ 5A - 4 B &=& 3\end{eqnarray} with the solution \(A = \frac{11}{9}\) and \(B= \frac{7}{9}\). The representation with proper rational functions is \[\frac{2x+3}{(x-4)(x+5)}=\frac{11}{9}\frac{1}{(x-4)}+\frac{7}{9}\frac{1}{(x+5)} \] The integral of the type \(\int \frac{ax+b}{(x-\lambda_1)(x-\lambda_2)}\,\mathrm{d} x.\) is no longer mystic.

With the help of partial fraction decomposition, this integral can now be calculated in the following manner \begin{eqnarray}\int \frac{ax+b}{(x-\lambda_1)(x-\lambda_2)}\mathrm{d} x &=& \int\frac{A}{(x-\lambda_1)}+\frac{B}{(x-\lambda_2)}\mathrm{d} x \\ &=&A\int\frac{1}{(x-\lambda_1)}\mathrm{d} x +B\int\frac{1}{(x-\lambda_2)}\mathrm{d} x \\ & = & A\ln(|x-\lambda_1|) + B\ln(|x-\lambda_2|).\end{eqnarray}

Example 3

Determine the antiderivative for \(\frac{2x+3}{(x-4)(x+5)}\), i.e. \(\int\frac{2x+3}{(x-4)(x+5)}\,\mathrm{d} x.\)

From the above example we already know: \[\int\frac{2x+3}{(x-4)(x+5)}\,\mathrm{d} x = \int\frac{11}{9}\frac{1}{(x-4)}+\frac{7}{9}\frac{1}{(x+5)}\, \mathrm{d} x.\]

Using the idea explained above immediately follow: \[\int\frac{11}{9}\frac{1}{(x-4)}+\frac{7}{9}\frac{1}{(x+5)} \,\mathrm{d} x = \frac{11}{9}\int\frac{1}{(x-4)} \mathrm{d} x + \frac{7}{9}\int\frac{1}{(x+5)} \,\mathrm{d} x= \frac{11}{9}\ln(|(x-4)|)+\frac{7}{9}\ln(|(x+5)|).\] So is the result \[\int\frac{2x+3}{(x-4)(x+5)}\,\mathrm{d} x=\frac{11}{9}\ln(|(x-4)|)+\frac{7}{9}\ln(|(x+5)|).\]

Case 2: \(Q(x)=(x-\lambda)^2\).

In this case \(R\) has the representation \(R(x) = \frac{ax+b}{(x-\lambda)^2}\) and the ansatz \[\frac{ax+b}{(x-\lambda)^2} = \frac{A}{(x-\lambda)}+\frac{B}{(x-\lambda)^2}\] is used.

By multiplying the equation with \((x-\lambda)^2\) we get \[ax+b = A(x-\lambda) + B.\] Again equating the coefficients leads us to a system of linear equations in \(A=a\) and \(B=b+A\lambda=b+a\lambda.\)

So we have \[\int \frac{ax+b}{(x-\lambda)^2}\,\mathrm{d}x = \int \frac{a}{(x-\lambda)}+\frac{b+a\lambda}{(x-\lambda)^2} \mathrm{d}x =a\ln(|x-\lambda|)-\frac{b+a\lambda}{(x-\lambda)}+c,\,c\in\mathbb{R}. \]

3. case \(Q(x)=x^2+mx+n\) without real zeros.

In this case \(R\) has the representation \(R(x) = \frac{ax+b}{x^2+mx+n}\) and the representation can not be simplified.

Only the special case \(R(x) = \frac{2x+m}{x^2+mx+n}\) is now considered.

In this case we have \[\int \frac{2x+m}{x^2+mx+n}\,\mathrm{d}x = \ln(|x^2+mx+n|)+c, \quad c\in\mathbb{R}. \]

Another special case is \(R(x) = \frac{1}{x^2+1}\) with \[\int \frac{1}{x^2+1} \, \mathrm{d} x = \arctan(x) +c,\quad c\in \mathbb{R}.\]

Integration by Parts

The derivative of a product of two continuously differentiable functions \(f\) and \(g\) is \[(f(x)\cdot g(x))' = f'(x)\cdot g(x)+f(x)\cdot g'(x),\quad x\in(a,b).\]

This leads us to the following theorem:

Proof

Example

Solve the integral \(\displaystyle\int_{0}^{\pi}x\sin x\,dx\).

Solution. Set \(f'(x)=\sin x\) and \(g(x) = x\). Then \(f(x)=-\cos x\) and \(g'(x) = 1\) and the integration by parts gives \[\int_{0}^{\pi}x\sin x\,dx = -\pi\cos\pi - 0 - \int_{0}^{\pi}(-\cos x)\,dx\] \[=\pi+\left(\sin x\right)\Bigg|_{x=0}^{\pi} = \pi.\]

Notice that had we chosen \(f\) and \(g\) the other way around this would have led to an even more complicated integral.

Integration by Substitution

Proof.

Example 1

Find the value of the integral \(\displaystyle\int_{0}^{\pi^2}\sin\sqrt{x}\,dx\).

Solution. Making the substitution \(x=t^{2}\) when \(t\ge0\) we have \(dx=2t\,dt\). Solving the limits from the inverse formula i.e. \(t=\sqrt{x}\) we find that \(t(0)=0\) and \(t(\pi^{2})=\pi\). Hence \[\int_{0}^{\pi^{2}}\sin\sqrt{x}\,dx = \int_{0}^{\pi^{2}}2t\sin t\,dt = 2\int_{0}^{\pi}t\sin t\,dt = 2\pi.\]

Here the latter integral was solved applying integration by parts in the previous example.

Example 2

Find the antiderivative of \(\displaystyle\frac{1}{\sqrt{x}(1+x)}\).

Solution. Substituting \(x=t^{2}\), \(t>0\) or \(t=\sqrt{x}\) gives \[\int\frac{dx}{\sqrt{x}(1+x)} = \int\frac{2t}{t(1+t^{2})}\,dt = 2\arctan t + C = 2\arctan\sqrt{x} + C.\]

The order of a differential equation is the highest order of the derivatives appearing in the equation.

For example, the order of the differential equation \( y' + 3y = \sin(x)\) is 1. The order of the differential equation \( y'' + 5y' -6y = e^x \) is 2.

Here the variable of the function \(y\) is not visible; the equation is considered to determine \(y\) implicitly.

A differential equation of order n is of the form 

\( \begin{equation} \label{dydef} F(x, y(x), y'(x),\ldots , y^{(n)}(x)) = 0 \end{equation} \)

The solution to an ODE is an n times differentiable function \(y(x)\) satisfying the above equation for all \( x \in I, \) where \(I\) is an open interval in the real axis.

Typically the solution is not unique and there can be an infinite number of solutions. Consider the equation \( xy^2 + y' = 0. \) The equation has the solutions

• \( y_0(x) = 0,\enspace x \in \mathbb{R} \)
  
• \( y_1(x) = 2/x^2,\enspace x>0 \)
  
• \( y_2(x) = 2/x^2,\enspace x<0 \)
  
• \( y_3(x) = 2/(x^2 + 3),\enspace x \in \mathbb{R} \)
  

Here \( y_1\), \( y_2\) and \( y_3 \) are called particular solutions. The general solution is \( y(x) = 2/(x^2 + C),\> C \in \mathbb{R}. \) Particular solutions can be derived from the general solution by assigning the parameter \(c\) to some value. Solutions that cannot be derived from the general solution are called special solutions.

Differential equations do not necessarily have any solutions at all. For example, the first order differential equation \( \sin(y' + y) = 2 \) does not have any solutions. If a first order equation can be written in normal form \( y' = f(x,y) \), where \(f\) is continuous, then a solution exists.

Constants in the general solution can be assigned to some values if the solution is required to satisfy additional properties. We may for example demand that the solution equals \( y_0 \) at \( x_0 \) by setting an initial condition \( y(x_0) = y_0. \) With first order equations, only one condition is (usually) needed to make the solution unique. With second order equations, we need two conditions, respectively. In this case, the initial condition is of the form

\( \left\{ \begin{array} yy(x_0) = y_0 \\ y'(x_0) = y_1 \end{array} \right. \)

In general, for an equation of order n, we need n extra conditions to make the solution unique. A differential equation and its set initial conditions are jointly referred as an initial value problem.

Example 1.

We saw above that the the general solution to the differential equation \( xy^2 + y' = 0 \) is \( y(x) = 2/(x^2 + C).\) Therefore the solution to the initial value problem

\( \left\{\begin{align} xy^2 + y' = 0 \\ y(0) = 1 \end{align} \right. \)

is \( y(x) = 2/(x^2 + 2).\)