Determine a particular antiderivative

Pick the curve through a point

The top graph shows \(f(x) = 3\sqrt{x}\). The faint curves below are antiderivatives of \(f\). They are copies of one curve at different heights, because they all have the same slope at each \(x\).

The ring marks the point \((4, 20)\), and you want the antiderivative whose curve passes through it.

Move the slider. The teal curve is \(\cb{2x^{3/2}} + C\), and the red segment is the gap between the teal curve and the ring at \(x = 4\).

At \(C = 0\) the teal curve has height 16 at \(x = 4\), so the gap is 4. The gap closes when \(C = 4\), and only this value works, because the ring picks one curve out of the family.

One point fixes one constant

Move the point anywhere. Press the button, and the teal curve moves up or down until it passes through the point.

A vertical line at the x of the point crosses each antiderivative at a different height, so only one of them passes through the point.

The red gap at that \(x\) is the distance from the teal curve to the point, so it is the amount \(C\) must change.

One point fixes one constant.

Notation

1. Write the general antiderivative, where \(G\) is any one antiderivative of \(f\) and \(C\) is the constant.

   \[F(x) = \cb{G(x)} + C\]

2. A condition such as \(F(a) = b\) is called an initial condition, because it gives one value of \(F\). The curve must pass through the point \((a, b)\), so substitute \(a\) for \(x\) and \(b\) for \(F(x)\).

   \[G(a) + C = \cy{b}\]

3. Solve for \(C\). The red gap shows how far the copy with \(C = 0\) is from the point.

   \[C = b - G(a)\]

4. Put this value of C back into F. The result is the particular antiderivative, the one curve that meets the condition.

5. Each antiderivative step adds one constant, so if you find an antiderivative twice, you need two conditions.

When it fails

The condition must give a value of the antiderivative. Suppose instead that you are given the slope at \(a\), shown in orange.

Slide \(C\). Every copy has slope \(f(a)\) at \(a\), so the orange tangent keeps its slope while the copy moves. A slope cannot pick out one copy, but a value can.

Example 1: One condition

Find the function \(F\) with \(F'(x) = 4x - \dfrac{1}{x^2}\) for \(x > 0\) and \(F(1) = 5\).

1. Find the general antiderivative. The term \(4x\) gives \(2x^2\). Write \(-\tfrac{1}{x^2}\) as \(-x^{-2}\), raise the exponent to \(-1\) and divide by \(-1\), which gives \(x^{-1}\).

   \[F(x) = 2x^2 + \frac{1}{x} + C\]

   The orange segments show the slopes \(f\) gives, and they fit every copy.

2. Substitute \(x = 1\) and \(F(1) = 5\).

   \[2 + 1 + C = 5\]

3. Solve for \(C\). The copy moves up by 2, and the red gap closes.

   \[C = 2\]

4. Put \(C = 2\) back. As a check, \(F(1) = 2 + 1 + 2 = 5\).

   \[F(x) = \cb{2x^2 + \frac{1}{x} + 2}\]

Example 2: Two conditions in a row

A sled has acceleration \(a(t) = 12t\) meters per second squared. At \(t = 0\), its velocity is \(-3\) meters per second and its position is 5 meters. Find its position \(s(t)\).

1. Velocity is an antiderivative of acceleration. Use the velocity at \(t = 0\) to find the constant.

   \[\begin{aligned} v(t) &= 6t^2 + C \\ v(0) = -3 &\;\Rightarrow\; C = -3 \end{aligned}\]

2. Position is an antiderivative of velocity. Add a new constant \(D\), because this is a second antiderivative step.

   \[s(t) = 2t^3 - 3t + D\]

3. A tempting wrong move: Use the velocity \(-3\) to find \(D\).

   \[s(0) = -3 \;\Rightarrow\; D = -3\]

   The number \(-3\) is the velocity at \(t = 0\), not the position. The sled starts at 5 meters, but the red particle starts at \(-3\).

4. Use the position at \(t = 0\), which is 5.

   \[s(0) = D = 5\]

5. Write the particular antiderivative. As a check, \(s'(0) = -3\) and \(s(0) = 5\).

   \[s(t) = \cb{2t^3 - 3t + 5}\]

Practice

Find the function \(g\) with \(g'(x) = 9x^2 - 4\) and \(g(2) = 10\).

Slide \(C\) until the teal curve passes through the ring at \((2, 10)\). The ring lies below the curve with \(C = 0\), so \(C\) is negative, and its size is the red gap at \(x = 2\).

Questions

The top graph shows \(f(x) = 5 - 2x\). The lower graph shows copies of the antiderivative \(F(x) = 5x - x^2\), raised or lowered by \(C\). The ring marks the point \((2, 9)\).

1. The copy with \(C = 0\) has height 6 at \(x = 2\). What is \(C\) for the antiderivative through the ring?

   • 3
   • 9
   • 6
   • −3
   Answer

   3

2. How many antiderivatives of \(f\) pass through the ring?

   • Exactly one
   • None
   • Two
   • Infinitely many
   Answer

   Exactly one

3. The function \(f\) has \(f'(x) = 6x^2 - 2\) and \(f(-1) = 4\). What is \(f(2)\)?

   • 16
   • 12
   • 8
   • 4
   Answer

   16

Summary

You write the general antiderivative, substitute the given point, solve for C, and put C back.

This works because the antiderivatives of a function are shifted copies of one curve, and only one copy passes through a given point.

You use it when a problem gives a derivative and a value, with one value for each antiderivative step.

This is one of five skills in topic 6.8 of Which & Why: AP Calculus.