We find the function y = f(x) corresponding to the y = f'(x) function depicted below, subject to the initial condition that f(0) = 12.

• Recalling that the graph of y = f'(x) gives us the slopes of the y = f(x) graph, we see that for the first interval from x = 0 to x = 10 this slope is increasing and the graph of f(x) is therefore concave upward, and with a positive slope due to the positive value of f'(x).
• Over the next interval, from x = 2 and to x = 30, f(x) becomes concave downward though maintaining a positive slope until x = 30, at which point the slope will be zero.
• The slope then continues decreasing between x = 30 and x = 40, while its slope remains negative over this interval. f(x) will therefore be decreasing and concave downward over this interval.
• Over the final interval, from x = 40 to x = 60, the slope increases so that the graph becomes concave upward. The slope of the graph however remains negative until x = 60, at which point the slope will be zero.

By the Fundamental Theorem, which tells us that the change `df over an interval is found by integrating f'(x) over the interval, the changes in the value of f(x) are represented by the areas under the graph of f'(x).

• These areas are easily found and are depicted in the figure above.

From these areas we would conclude, as shown below, that

• f(10) = f(0) + 40 = 12 + 40 = 52,
• f(30) = f(10) + 50 = 52 + 50 = 102,
• f(40) = f(30) + -15 = 102 - 15 = 87 and
• f(60) = f(40) + -30 = 87 - 30 = 57.

The figure below shows a table of values of f(x) vs. x, with the behavior of f(x) indicated for each x interval.

• We note that (30, 102) and (60, 57) will be critical points, where f'(x) = 0.
• We note also that at x = 10 and at x = 40, the slope of f'(x) changed sign so that the concavity of f(x) changed, indicating that these points are inflection points.
• The graph shown below corresponds to these behaviors.

The graph of a function y = f''(x) is depicted in the graph below.

From this graph of f'' we can determine the behavior of its antiderivative f'(x).

• Since the values of f'' are decreasing for negative values of x, it follows that f' will be concave downward for negative x values.
• Since f'' is positive for negative values of x, it follows that f' will be an increasing function for negative x values.
• Since the values of f'' are increasing for positive values of x, it follows that f' will be concave up for positive x values.
• Since f'' is negative for positive values of x, it follows that f' will be a decreasing function for positive x values.
• Since f'' is 0 at x = 0, we conclude that f' has a critical point at x = 0.

The graph of f'(x) is shown below, wth the specified value f'(0) = 0.

We now use the above graph of f' to determine the behavior of the function f.

• Except for a critical point of f at x = 0, where f'(x) = 0, we see that f'(x) is negative, so that f will be decreasing.
• For negative values of x, f'(x) is increasing, so f must be concave upward.
• For positive values of x, f'(x) is decreasing so f must be concave downward.

The corresponding behavior of the graph of f(x) is shown below.

The figure below depicts the graphs of two functions f(x) and g(x), which intersect one another at the x axis.

• We wish to find graphs of antiderivative functions F and G of f and g, with F(0) = G(0) = 0.
• Since both the graphs of f(x) and g(x) are strictly increasing, the graphs of their antiderivative functions F and G will be always concave upward.
• To the left of the point where the two graphs meet, both f and g are negative so the both the antiderivative functions F and G will be decreasing.
• To the right of the point where the two graphs meet, both f and g are positive so both the antiderivative functions F and G will be increasing.
• Both F and G will have critical values at the x value where f and g are zero--i.e., at the intersection point.
• Since the magnitude of g(x) is always greater than that of f(x), we see that g(x) will always have the steeper graph.

The figure below shows the corresponding behavior of the graphs of F(x) and G(x).

• Both graphs start at the origin, since F(0) = G(0) = 0.
• Both graphs begin with a negative slope and remain concave upward.
• Since as we observed earlier the graph of G(x) is always steeper than that of F(x), we see that G(x) immediately becomes less that F(x) and, until the common critical point, G(x) continues to decrease faster than F(x).
• To the right of the critical point, the steeper graph of G(x) will increase faster than F(x), so that G(x) will at some point overtake F(x).

To find an antiderivative function of 3 sin(5t), which we write as the integral in the first line below, we recall that

• sine functions are obtained by taking the derivatives of cosine functions
• the argument of the cosine function will be the same as the argument of the derivative sine function (i.e., since the argument of the sine function is 5t, the argument of the cosine function will also be 5t)
• there is such a thing as the chain rule.

We know that the derivative of cos(5t) is - 5 sin(5t).

• It would follow that the derivative of -1/5 cos(5t) is equal to sin(5t).
• From this it follows that to get 3 sin(5t), we need to take the derivative of 3 * (-1/5 cos(5t)).

Knowing that we get the derivatives of sines from cosine functions, we make an educated guess that the antiderivative function will be some multiple of cos(5t).

• We therefore set up a trial solution F(t) = c cos(5t), where c is a constant to be determined.
• We easily see that the derivative of F(t) is F'(t) = - 5 c sin(5t).
• Since we want the derivative to equal 3 sin(5t), we set the two expressions for F'(t) equal, obtaining the equation in the next-to-last line below.
• The equation is easily solved; we determine that c = -3/5.
• We then proceed to substitute this expression into the expression for F(t), obtaining F(t) = -3/5 cos(5t) as before.

To find an antiderivative of e^(7t + 3), we observe that the derivative of e^(7t + c) is 7 e^(7t + 3).

• We conclude that the function whose derivative is e^(7t + 3) is 1/7 e^(7t + 3).

We might try to find an antiderivative for e^(x^2).

We would probably be tempted to try a function of the form c e^(x^2).

• The derivative of c e^(x^2) is 2x c e^(x^2).
• If we were to set this expression equal to e^(x^2), we would find that c = 1 / (2x). This is not a constant number, since x is a variable.
• We might decide to try using this value of c anyway, especially if we have nothing better to try. But we aren't going to hold out a great hope of success.
• It's a good thing we didn't hold out much hope of success, since by the product rule the derivative of 1 / (2x) e^(x^2) has two terms, only one of which is e^(x^2).

On the other hand, we can easily enough find an antiderivative of x^2 e^(x^3), as shown below.

• We see that the x^2 occurs as a result of the chain rule.