Building an antiderivative

Building an Antiderivative

We begin with the function f defined as follows:

f(x) = -1 if x <= -1

x if -1 < x <= 1

2 - x² if 1 < x <= 2

-2 if x > 2

Here is the graph, with the gray lines marking the intervals on which the function is defined by different expressions. Note that the expressions are chosen so that the resulting function is continuous. (For example, its limit as x approaches -1 from the left is equal to its limit from the right, i.e., y = -1.)

Thus, according the (first part of) the Fundamental Theorem of Calculus, the function F defined as the integral from 1 to x of f (I wish it were possible here to put an integral sign into a browser) has the property that its derivative is f. Thus, we have:

F(x) = -x + C₁ if x <= -1

.5x² + C₂ if -1 < x <= 1

2x - (1/3)x³ + C₃ if 1 < x <= 2

-2x + C₄ if x > 2

for some constants C₁, C₂, C₃, and C₄.

Thus, the graph of F consists of pieces on each of the intervals on which f is defined, as shown to the right. But since we do not yet know the values of the constants C₁ to C₄, we don't yet know which piece is the correct one on each interval -- choosing different values of a constant moves the corresponding piece of the curve up and down. We will have determined F when we have found these constants.

We first choose C₂. Because F(x) is defined as the integral of f from 1 to x, it follows that F(1) is the integral of f from 1 to 1; so F(1) = 0 -- the integral of any function over an interval of zero length is 0. Thus, we must choose C₂ so that the value of .5x² + C₂ is 0 when x=1. This forces
C₂ = -.5(1)² = -.5,
and we have determined one part of the antiderivative.

Next we use that fact that F is continuous (because it has a derivative) to choose C₃. This constant is chosen so that the limit of F as x approaches 1 from above is 0 (i.e., F(1), as we noted while finding C₂), so:
2(1) - (1/3)(1)³ + C₃ = 0 C₃ = -5/3

Similarly, we choose C₁ so that the piece of the graph for x-values less than -1 joins with the piece of the graph for x-values between -1 and 1:
-(-1) + C₁ = .5(-1)² - .5 C₁ = -1 And finally we choose C₄ so that two pieces match at x=2:
2(2) - (1/3)(2)³ - (5/3) = -2(2) + C₄ C₄ = 11/3

Finally, we note that this F is only one of the antiderivatives of f, namely the one whose value at x = 1 is 0. If we add any constant to F throughout, we get a different antiderivative of f. The graph in green shows the result of adding .5 to F, resulting in an antiderivative of f that has value 0 at x = 0.

f(x) =	-1	if x <= -1
	x	if -1 < x <= 1
	2 - x²	if 1 < x <= 2
	-2	if x > 2

F(x) =	-x + C₁	if x <= -1
	.5x² + C₂	if -1 < x <= 1
	2x - (1/3)x³ + C₃	if 1 < x <= 2
	-2x + C₄	if x > 2

Thus, the graph of F consists of pieces on each of the intervals on which f is defined, as shown to the right. But since we do not yet know the values of the constants C₁ to C₄, we don't yet know which piece is the correct one on each interval -- choosing different values of a constant moves the corresponding piece of the curve up and down. We will have determined F when we have found these constants.
We first choose C₂. Because F(x) is defined as the integral of f from 1 to x, it follows that F(1) is the integral of f from 1 to 1; so F(1) = 0 -- the integral of any function over an interval of zero length is 0. Thus, we must choose C₂ so that the value of .5x² + C₂ is 0 when x=1. This forces C₂ = -.5(1)² = -.5, and we have determined one part of the antiderivative.
Next we use that fact that F is continuous (because it has a derivative) to choose C₃. This constant is chosen so that the limit of F as x approaches 1 from above is 0 (i.e., F(1), as we noted while finding C₂), so: 2(1) - (1/3)(1)³ + C₃ = 0 C₃ = -5/3
Similarly, we choose C₁ so that the piece of the graph for x-values less than -1 joins with the piece of the graph for x-values between -1 and 1: -(-1) + C₁ = .5(-1)² - .5 C₁ = -1 And finally we choose C₄ so that two pieces match at x=2: 2(2) - (1/3)(2)³ - (5/3) = -2(2) + C₄ C₄ = 11/3
Finally, we note that this F is only one of the antiderivatives of f, namely the one whose value at x = 1 is 0. If we add any constant to F throughout, we get a different antiderivative of f. The graph in green shows the result of adding .5 to F, resulting in an antiderivative of f that has value 0 at x = 0.