Primitive functions

Reversing differentiation: primitives of powers of x including negative and fractional powers, the arbitrary constant, finding a particular primitive from an initial condition, and working back from a second derivative.

Reversing differentiation

$F(x)$ is a primitive of $f(x)$ if $F'(x) = f(x)$. For example, $x^3$ is a primitive of $3x^2$. So are $x^3 + 5$ and $x^3 - 2$, because a constant differentiates to $0$.

Primitive of a power (n ≠ −1)

$$\text{If } f'(x) = x^n, \text{ then } f(x) = \frac{x^{n+1}}{n + 1} + C$$

Increase the power by one, then divide by the new power. The rule fails for $n = -1$, which needs logarithms (a later chapter).

A family of primitives of 2x

Every curve y = x² + C has gradient 2x. Changing C moves the curve up or down without changing any gradient.

Try this

Which value of $C$ makes the curve pass through $(1, 3)$?

Example 1●

Find the primitive of $6x^2 - 4x + 3$.

$2x^3 - 2x^2 + 3x + C$

Example 2: negative and fractional powers●●

Find primitives of $\dfrac{1}{x^3}$ and $\sqrt{x}$.

$\dfrac{1}{x^3} = x^{-3}$, with primitive $\dfrac{x^{-2}}{-2} + C = -\dfrac{1}{2x^2} + C$.

$\sqrt{x} = x^{\frac12}$, with primitive $\dfrac{x^{\frac32}}{\frac32} + C = \dfrac23 x^{\frac32} + C$.

Using an initial condition

Example 3●●

Given $f'(x) = 4x - 3$ and $f(2) = 5$, find $f(x)$.

$f(x) = 2x^2 - 3x + C$.

$f(2) = 8 - 6 + C = 5$, so $C = 3$ and $f(x) = 2x^2 - 3x + 3$.

Example 4: from the second derivative●●●

Given $f''(x) = 6x$, $f'(1) = 2$ and $f(0) = 4$, find $f(x)$.

$f'(x) = 3x^2 + C_1$, and $f'(1) = 2$ gives $C_1 = -1$.

$f(x) = x^3 - x + C_2$, and $f(0) = 4$ gives $C_2 = 4$. So $f(x) = x^3 - x + 4$.

Flashcards

Primitive of $x^n$

$\dfrac{x^{n+1}}{n + 1} + C$, for $n \ne -1$.

Why $+,C$?

Constants differentiate to $0$, so they can't be recovered.

Initial condition

Substitute the given point to find $C$.

Practice

Given $f'(x) = 4x - 3$ and $f(2) = 5$, find $f(0)$.
Given $f''(x) = 6x$, $f'(1) = 2$ and $f(0) = 4$, find $f(2)$.
A primitive of $x^4$ is:
The primitive of $\dfrac{1}{x^2}$ is:
The primitive of $\sqrt{x}$ is:
Given $f'(x) = 2x$ and $f(1) = 4$, find $f(3)$.
Why is there a $+\,C$ in every primitive?
The rule $\dfrac{x^{n+1}}{n + 1} + C$ does not work when:
For the family $y = x^2 + C$, which $C$ gives a curve through $(1, 3)$?