From our previous interpretations of differential equations using position or cost functions, we can suggest the following affirmative result: any differential equation of the form F’(x) = P(x) where P is a continuous function has a solution. For a position interpretation, just imagine you control the movement of an object by adjusting its velocity so that at any time, t, its velocity is P(t). Then the resulting position function for the object solves the differential equation. Or for an economics interpretation, imagine producing some good stuff where the marginal cost of producing x units is P(x). Then the resulting accumulated cost function for production solves the differential equation.
The motion and cost interpretations can be visualized with transformation figures. In this section we will use an area interpretation to produce a function that solves a differential equation. The advantage of focusing on area is that an area interpretation comes with its own inherent visualization. We will assume in this discussion a naive understanding of the concept of area as a numerical measure of a planar region without any deeper investigation of its meaning. We will explore a deeper meaning for area in Chapter V.
Example IV.G.1. [Constant velocity] Consider the differential equation: v(t) = s'(t) = 4 with initial condition s(0) = 0 where s denotes the position of an object moving on a straight coordinate line and v denotes its velocity. In other words, the object is moving at a constant velocity and started at the point with coordinate 0.
Using the notation of indefinite integrals, we can solve the differential
equation symbolically. Since ò v(t) dt
= ò 4 dt =
4t + C and s'(t) = v(t), we have that


Example IV.G.2. [ Constant Acceleration]. We continue to look
at velocity and position. This time use the differential equation s'(t)
= v(t) = 3t with s(0)=0. Now our object has an increasing
velocity and a constant acceleration. Again we can solve precisely for
s(t). Since òv(t)dt = ò3t
dt = (3/2)t^{ 2} + C, we have (from Theorem 4.2)
that s(t) = (3/2)t^{ 2} + C. But s(0) = 0 implies
that C = 0, so s(t) = (3/2)t^{ 2} .
Now in the graph of v of Figure IV.G.ii we observe that the shaded region (a triangle) which is enclosed by the Xaxis, the lines X = 0 and X = t, and v = 3x has area equal to one half the base times the altitude of the triangle, i.e., (1/2)(t)(3t) = (3/2)t^{ 2}. So the area again is precisely the value for s(t). 

Example IV.G.3. [Solving a differential
equation using an area function.]Consider a moving object with a positive
velocity decreasing to 0. Perhaps we have observed that the object is slowing
down (to a stop) but always continues to move. There are several possible
equations that might model this situation. For this example let's consider
v(t) = s'(t) = 1/(t+1) where t > 1 and
s(0)=0. The graph of v is in Figure IV.G.iii.
Our problem is to describe at least one function that solves this differential equation for t > 0. Following the pattern suggested by the two previous examples, for t > 0 we let F(t) denote the area of the region enclosed by the X  axis, X = 0, X = t, and Y = 1/(x+1). Notice that F(0)=0 because the "region" in this case is a line segment and its area is 0. Recall that the work in Section IV.F demonstrated that estimates for the value of the position function, s(t), using Euler's method would also serve as estimates for F(t). Thus, it should seem reasonable that s(t) = F(t) and that 

Justification that F'(a) = 1/(a+1) for any a > 0 : We begin by interpreting the expression F(a+h)  F(a) when h > 0. Consider the graph in Figure IV.G.iv. On this graph the region enclosed by the Xaxis, X = a, X = a+h, and the graph of Y = 1/(X+1) has area F(a+h)  F(a). 

Examining this region more carefully in Figure IV.G.v
we see that 1/(a+h+1)^{ . }h < F(a+h)
 F(a) < 1/(a+1)^{ . }h because the region is enclosed
within a large rectangle with area h/(a+1) and encloses a
smaller rectangle with area h/(a+h+1).
Now to find F'(a) we consider [F(a+h)  F(a)] / h as h > 0. From the previous work we now see that for h > 0 

14  21. For the functions v given in problems 18, sketch a tangent field for the differential equation S'(t) = v(t). Sketch an integral curve on the field passing through the point (0,0). Use this integral curve to estimate the area of the region enclosed by the X? axis, X = 0, X = 2, and the graph of Y = v(x). Discuss the relation of the integral curve to the function F(t) which denotes the area of the region enclosed by the Xaxis, X = 0, X = t, and Y= v(x).
22. Complete the discussion for F(t) being a solution to the differential equation v(t) = s'(t) = 1/(t+1) where t > 1 and s(0) = 0 by examining the situation with h < 0 and showing that
23. Suppose P(x) has a graph as given in Figure IV.G.vi and
F(t) is defined as the area of the region enclosed by the Xaxis,
the Yaxis, the line X= t, and the graph of Y=P(x).
a) Estimate F(2.1), F(2.1)  F(2), F'(2). b) Estimate F'(2), F'(3), F'(1). c) For what value of t do you think F'(t) is largest on the interval [0,4]? d) For what value of t do you think F'(t) is smallest on the interval [0,4]? e) Explain why F(t) is an increasing function on the interval [0,4]. f) Discuss the concavity of the graph of F. 
