We recall that the Second Fundamental Theorem essentially says that the function F(x) defined by the integral in the figure below, with a constant lower limit and x as upper limit, is an antiderivative of the function f(x) being integrated (note that f(x) is integrated with the dummy variable t).

• The geometric interpretation of this integral identifies it with the area of the graph in the figure.
• As x moves across the t axis, the area changes.
• The changing area is a function of x, and this function is an antiderivative of the y vs. t function f whose graph defines the curve.

It follows that the x derivative of F(x) is just f(x).

As an example, the integral in the figure below is seen by the Second Fundamental Theorem to be an antiderivative of the function f(t) = e^(t^4 - 3 t^2 + 2 t) which is being integrated.

• The derivative of the integral is therefore the derivative of an antiderivative of f(x), which is just f(x).

In the figure below we construct a graph of F(x), as defined by the indicated integral.

We begin with the graph of the sine function being integrated, depicting a complete cycle from t = 0 to t = 2 `pi.

• Since F(x) is an antiderivative of the sine function, the value of the sine function will give us the slope of the graph of F(x).
• For any x, F(x) is the accumulated area of the graph of the sine function, from t = 0 to t = x.
• We see that F(0) = 0, since the area under the graph from 0 to 0 is 0.
• Since the sine function is positive and increasing for the first 1/4 of its period, the slope of the graph of F(x) will be increasing and concave upward from x = 0 to x = `pi / 2.
• Since the sine function is positive and decreasing between `pi/2 and `pi, F(x) will be increasing and concave downward over this range of values.
• The inflection point between the two concavities will occur at x = `pi/2.
• The sine function then becomes negative, so that F(x) will decrease; the sine function is first decreasing so that F(x) is concave downward, then increasing so that F(x) becomes concave upward; the inflection point occurs at x = 3 `pi / 2.
• A critical point occurs at `pi, where the sine function is 0 and F(x) therefore has derivative 0.

We define erf(x) by the integral shown in the first line of the figure below.

Though we cannot evaluate this integral for most values of x, the Second Fundamental Theorem allows us to determine its derivative at any x.

We note that the derivative of erf(x) is, by the Second Fundamental Theorem, e^-(x^2).

Using this knowledge we can find the derivative of x^2 erf(x):

• We first use the product rule, then we apply our knowledge of the derivative of erf(x) to see that erf'(x^3) = e^-(x^3)^2.
• We obtain the expression in the last line, which can be slightly simplified by squaring the x^3 in the exponent of the exponential function.

To determine the time at which an object dropped from rest at an altitude of 400 feet strikes the ground, we write the differential equations with initial conditions that describe the situation.

• Since the acceleration is constant and equal to -g, and since acceleration is defined as dv / dt, we write the differential equation dv / dt = -g.
• Since the initial velocity of an object dropped from rest is 0, we write the initial condition v(0) = 0.
• Since velocity is the derivative of position, we write ds / dt = v.
• Since the initial position of the object is 400 feet above the ground, we write s(0) = 400.

We proceed to solve the system of equations.

• Since the general antiderivative of -g is -gt + c, we see that v(t) = -g t + c.
• Applying the initial condition v(0) = 0, we see that c = 0.
• We conclude that v(t) = - g t.

Using this velocity function in the differential equation ds / dt = v, we solve this second equation.

• We see that ds / dt = - g t.
• The general antiderivative of - g t is - 1/2 g t^2 + c; this is our function s(t), as indicated in the second line below.
• Applying the initial condition s(0) = 0 we see that c = 400 and s(t) = - 1/2 g t^2 + 400.

We now determine the clock time t at which the object reaches the ground.

• The object reaches the ground when its position s(t) = 0.
• Using the function s(t), we obtain the equation in the second line from the bottom of the figure below.
• This equation is easily solved to find t = `sqrt( 800 / g).

Suppose that we wish to find the value of g for which the object would reach the ground twice as fast.

We first note that the time required is t = `sqrt(800/g).

• By changing g we change the number of which were taking the square root.
• To make a square root half as great, we have to make the number of which are taking the square root 1/4 as great.
• If we make g four times as great, 800 / g will become 1/4 as great, and the square root will become 1/2 as great.
• Therefore to make t half as great, g must become four times as great.

We now solve the problem of when an object thrown upward with a vertical velocity of 10 ft / sec will return to the ground, assuming a gravitational acceleration of 5 ft / sec^2.

We find the position function s(t) corresponding to an acceleration of -5, an initial position 0, and initial velocity 10.

• Our differential equation for the velocity and the corresponding initial condition are indicated in the first line below.
• The velocity function is equal to the general antiderivative of -5, as indicated in the second line.
• In the third line we evaluate c using the initial condition.
• In the fourth line we write the resulting velocity function.
• We now set ds / dt equal to the velocity function, with the initial condition, as indicated in the fifth line.
• We solve this equation by the usual process to obtain s(t) = -2.5 t^2 + 10 t.

The motion of the object starts and ends at position s = 0.

• We can therefore determine the time at which the object strikes the ground by solving the equation s(t) = 0.
• Using the function s(t), we obtain the equation in the fifth line below.
• The equation is easily solved to find that the object is at ground level at times t = 0 and t = 4.
• The t = 0 time corresponds to the instant when the object was thrown upward, the t = 4 time to the instant when the object returns to the ground.

The object reaches its maximum altitude when its velocity reaches 0, at the instant between its upward and downward motions.

• Setting our function v(t) equal to zero we easily find that maximum altitude occurs at clock time t = 2.
• At this instant the altitude of the object is s(t) = s(2) = 10 feet.