Determination of the Constant of Integration

To determine the constant of integration, in addition to the given function to be integrated, we need extra data. Specifically, the extra data is the value of the integral function at a point. Let us illustrate this by means of examples.

Example 1

Find \(f\) if \(f'(x)=x^{3}-3x+1+e^{x}\) and \(f(0)=-1.\)

Solution 1

\[\begin{aligned} f(x)=\int f'(x)dx & =\int(x^{3}-3x+1+e^{x})dx\\ & =\frac{1}{4}x^{4}-\frac{3}{2}x^2+x+e^{x}+C\end{aligned}\] To determine \(C\), we use the condition \(f(0)=-1\): \[f(0)=0-0+0+\underbrace{e^{0}}_{=1}+C=-1\] \[\Rightarrow C=-2\] so \[f(x)=\frac{1}{4}x^{4}-\frac{3}{2}x^2+x+e^{x}-2.\]

Example 2

Find \(f\) if \(f^{\prime\prime}(x)=3\sin x+5x\), \(f(0)=1\), and \(f'(0)=0\).

Solution 2

\[\begin{aligned} f'(x) & =\int f^{\prime\prime}(x)dx\\ & =\int(3\sin x+5x)dx\\ & =-3\cos x+\frac{5}{2}x^{2}+C.\end{aligned}\] Because \(f'(0)=0\), we have \[-3\underbrace{\cos0}_{=1}+\frac{5}{2}(0)+C=0\] \[\Rightarrow C=3.\] Therefore, \[f'(x)=-3\cos x+\frac{5}{2}x^{2}+3,\] and \[\begin{aligned} f(x) & =\int f'(x)dx\\ & =\int\left(-3\cos x+\frac{5}{2}x^{2}+3\right)dx\\ & =-3\sin x+\frac{5}{6}x^{3}+3x+D.\end{aligned}\] To determine \(D\), we impose the condition \(f(0)=1\): \[f(0)=-3\underbrace{\sin0}_{=0}+\frac{5}{6}(0)+3(0)+D=1\] \[\Rightarrow D=1.\] Therefore \[f(x)=-3\sin x+\frac{5}{6}x^{3}+3x+1.\]

Example 3

Find \(f\), if \(f^{\prime\prime}(x)=1-x^{2}\), \(f(0)=2\), and \(f(1)=-1\).

Solution 3

\[\begin{aligned} f'(x) & =\int f^(x)dx\\ & =\int(1-x^{2})dx\\ & =x-\frac{x^{3}}{3}+C\end{aligned}\] and \[\begin{aligned} f(x) & =\int f'(x)dx\\ & =\int\left(x-\frac{x^{3}}{3}+C\right)dx\\ & =\frac{1}{2}x^{2}-\frac{1}{12}x^{4}+Cx+D.\end{aligned}\] To determine \(C\) and \(D\), we use the given conditions \(f(0)=2\) and \(f(1)=-1\): \[f(0)=0-0+0+D=2\Rightarrow D=2\] and \[f(1)=\frac{1}{2}(1^{2})-\frac{1}{12}(1^{4})+C(1)+2=-1\] \[\Rightarrow C=-\frac{41}{12}.\] Therefore, the required function is \[f(x)=\frac{1}{2}x^{2}-\frac{1}{12}x^{4}-\frac{41}{12}x+2.\]


Geometrical Significance of the Constant of Integration

Let us illustrate this by means of an example.

Suppose we want to determine the equation of the curve at every point of which the tangent line has the slope \(3x^{2}-2x\).

Because the slope of the tangent to a curve at a point is \(dy/dx\) or \(f'(x)\), we have by the hypothesis \[f'(x)=3x^{2}-2x.\] Therefore, by integration we have \[f(x)=\int(3x^{2}-2x)dx=x^{3}-x^{2}+C,\] where \(C\) is the constant of integration. This family of functions is shown in Figure 1 where we see curves for various choices of \(C\). There are an infinite number of them. All of them have the same value of \(f’\) (or \(dy/dx\)); that is, they have the same slope (or the same direction) for the same value of \(x\). Also, notice that the difference in the lengths of their \(y\)-values remains the same for all values of \(x\). Hence, all of these curves can be obtained by moving any one of them vertically up or down, the value of \(C\) in this case not affecting the slope of the curve.

Figure 1
Figure 1. Graphs of \(y=x^{3}-x^{2}+C\) for various values of \(C\).



If in this example, we impose the additional condition that the curve shall pass through the point \((1,2)\), then the coordinates of this point must satisfy \(y=x^{3}-x^{2}+C\), giving \[2=1-1+C,\quad\text{or}\quad C=2.\] Hence, the particular curve required is \[y=x^{3}-x^{2}+2.\]


Differential Equations and Initial-Value Problems

Many problems in mathematics, science, engineering, economics, etc. are posed mathematically in terms of an equation that involves derivatives of an unknown function. An equation that involves derivatives of an unknown function is called a differential equation. For example, \(f'(x)=3x^{3}-2x\) or \(f'(x)=-x/f(x)\) are two differential equations. They are often written as \[y’=3x^{2}-2x,\quad\text{and}\quad y’=-\frac{x}{y}.\]

We know that every function that satisfies \(f'(x)=3x^{2}-2x\) (or \(y’=x^{3}-x)\) is of the form \[y=f(x)=x^{3}-x^{2}+C.\] This is called the general solution of the given differential equation. The general solution of a differential equation involves an arbitrary constant (or constants). If we know the value of \(f\) for one value of \(x\), then we can determine a particular solution. For example, if we know \(f(1)=2\), then we obtain \[f(x)=x^{3}-x^{2}+2.\] This is called a particular solution and the extra information \(f(1)=2\) is called an initial condition solution. If this differential equation is written as \(y’=3x^{2}-2x\), it is common practice to write the given initial condition as \(y(1)=2\). In this section, we learned that the initial condition \(f(1)=2\) specifies only one curve in the family of curves that passes through the point \((1,2)\).

The problem of solving a differential equation subject to an initial condition is called an initial-value problem. Therefore \[y’=3x^{2}-2x,\quad y(1)=2\] is an initial value problem.

Recall that \(y’\) can be regarded as \(dy\) divided by \(dx\). Therefore, the problem \(y’=3x^{2}-2x\) is equivalent to \[\frac{dy}{dx}=3x^{2}-2x\] or \[dy=(3x^{2}-2x)dx.\] Now we can integrate each side \[\int dy=\int(3x^{2}-2x)dx\] \[\Rightarrow y+C_{1}=x^{3}-x^{2}+C_{2}\] We can combine \(C_{1}\) and \(C_{2}\) and simply write \[y=x^{3}-x^{2}+C,\] where \(C=C_{2}-C_{1}\). This technique, called separation of variables, can help us solve some differential equations. For instance, see the following example.

Example 4

Determine the equation of a curve such that the slope of the tangent to the curve at any point is the negative ratio of the \(x\)-value to the \(y\)-value (the abscissa to the ordinate) and it passes through the point \((3,4)\).

Solution 4

This is an initial-value problem which can be mathematically expressed as \[\frac{dy}{dx}=-\frac{x}{y},\quad y(3)=4.\] We can solve the differential equation \(dy/dx=-x/y\) by separation of variables. That is, we multiply both sides by \(ydx\) obtaining \[ydy=-xdx.\] Integrating each side gives us \[\frac{1}{2}y^{2}+C_{1}=-\frac{1}{2}x^{2}+C_{2}\] or \[x^{2}+y^{2}=C,\tag{i}\] where \(C=2(C_{2}-C_{1})\). Equation (i) is the general solution of \(y’=-x/y\). Several curves of the form \(x^{2}+y^{2}=C\) are shown in Figure 2.

If we impose the initial condition that the curve must pass through the point \((3,4)\) or \(y(3)=4\), we obtain \[\underbrace{x^{2}}_{3^{2}}+\underbrace{y^{2}}_{4^{2}}=C\Rightarrow C=25.\] Hence, the particular solution satisfying the differential equation and the initial condition is the equation of the circle \(x^{2}+y^{2}=25\).

Figure 2
Figure 2. The curves \(x^{2}+y^{2}=C\) for various values of \(C\). We identify the curve \(x^{2}+y^{2}=25\) as the one that passes through the point \((3,4)\).

Rectilinear Motion

If we know the position of a particle (as a function of time \(t\)) \(s(t)\), then the velocity of the particle \(v(t)\) is \[v(t)=s'(t)\] and its acceleration \(a(t)\) is \[a(t)=v'(t)=s^{\prime\prime}(t).\] Sometimes the reverse procedure is important. In other words, if we are given \(v(t)\) (or \(a(t)\)), we want to find \(s(t)\), the distance traveled since time \(t_{0}\) to where we are knowing the initial position (and velocity). This method is used for navigating submarines and spacecrafts.

Example 4

Find the equation describing the position of a particle if it moves in a straight line with constant acceleration, knowing the velocity and position of the particle at \(t=0\).

Solution 4

Because acceleration is a constant, say \(a_{0}\), we have \[v'(t)=a_{0}.\] Integrating \[v(t)=\int a_{0}dt=a_{0}t+C.\] To determine \(C\), suppose the velocity be \(v_{0}\) at \(t=0\); that is, let \[v(0)=v_{0}.\] Therefore \[v_{0}=a_{0}(0)+C\Rightarrow C=v_{0}\] and \[\begin{aligned} v(t) & =a_{0}t+C\\ & =a_{0}t+v_{0}.\end{aligned}\] Since \(s'(t)=v(t)\), we have \[s'(t)=a_{0}t+v_{0}\] Integrating gives us \[\begin{aligned} s(t) & =\int(a_{0}t+v_{0})dt\\ & =\frac{1}{2}a_{0}t^{2}+v_{0}t+C_1.\end{aligned}\] To determine \(C_1\), suppose the position be \(s_{0}\) at \(t=0\); that is, let \[s(0)=s_{0}.\] Therefore, \[s_{0}=0+0+C_1\Rightarrow C_1=s_{0}\] and \[s(t)=\frac{1}{2}a_{0}t^{2}+v_{0}t+s_{0}.\]

The force of gravitation exerted on an object is the weight of the object, \(mg\), where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity. For objects near the surface of the earth, we can assume that \(g\) is a constant (\(g=9.8\) m/s\(^{2}\) or \(g=32\) ft/s\(^{2}\), approximately).

  • If the positive direction is taken to be downward, then the gravitational acceleration is a positive constant; that is, \[a(t)=a_{0}=g.\] In this case, the equations describing the motion of a stone in vacuum falling from rest (\(v_{0}=0\)) are \[v(t)=gt,\] \[s(t)=\frac{1}{2}gt^{2}.\] assuming \(s=0\) at \(t=0\) (i.e. \(s_{0}=0\)). We may eliminate \(t\) between the above equations and find a relation between the position of the stone and its velocity at each moment: \[v(t)=\sqrt{2g\,s(t)}.\] If the object is initially thrown upward, we have \(v_{0}<0\).
  • If the positive direction of is taken to be upward, then \[a(t)=a_{0}=-g\] and if the object is initially thrown upward we have \(v_{0}>0\).