Today's lesson
Learning intentions
- Antidifferentiate $e^{kx}$ (and combinations like $(e^x - 5)^2$ after expanding)
- Antidifferentiate $\sin(kx+a)$ and $\cos(kx+a)$
- Use an initial condition (a point on the graph) to find the constant of integration $c$
- Use a stationary point condition to find an unknown constant in $f'(x)$
Work through Examples 6–9 below. Each one is fully worked — copy the method into your notes, then attempt the Cambridge questions at the end.
Key result — exponential
FORMULA
$\displaystyle\int e^{kx}\, dx = \tfrac{1}{k}e^{kx} + c, \quad k \neq 0$
📺 Concept: Why we need an initial condition. The family of curves $y=5e^{2x}+c$ all share the same gradient — the point $(0, 15)$ picks the one with $c = 10$.
📺 Example 6 walkthrough: Expand $(e^x-5)^2$ first, then antidifferentiate each term.
EXAMPLE 6
Find $\displaystyle\int (e^x - 5)^2\, dx$ by first expanding.
Expand the bracket (perfect square):
$(e^x - 5)^2 \;=\; (e^x)^2 - 2(5)(e^x) + 25 \;=\; e^{2x} - 10e^x + 25$
$(e^x - 5)^2 \;=\; (e^x)^2 - 2(5)(e^x) + 25 \;=\; e^{2x} - 10e^x + 25$
Now antidifferentiate term by term:
$\displaystyle\int e^{2x}\, dx = \tfrac{1}{2}e^{2x}$, $\displaystyle\int 10e^x\, dx = 10e^x$, $\displaystyle\int 25\, dx = 25x$
$\displaystyle\int e^{2x}\, dx = \tfrac{1}{2}e^{2x}$, $\displaystyle\int 10e^x\, dx = 10e^x$, $\displaystyle\int 25\, dx = 25x$
$\displaystyle\int (e^x - 5)^2\, dx \;=\; \tfrac{1}{2}e^{2x} - 10e^x + 25x + c$
SIMILAR EXAMPLE 6a
Find $\displaystyle\int 3e^{2x}\, dx$.
Use $\int e^{kx} dx = \tfrac{1}{k}e^{kx}$ with $k=2$:
$\displaystyle\int 3e^{2x}\, dx = 3 \cdot \tfrac{1}{2}e^{2x} + c$
$= \tfrac{3}{2}e^{2x} + c$
SIMILAR EXAMPLE 6b
Given $\dfrac{dy}{dx} = 10e^{2x}$ and the point $(0, 15)$ is on the graph of $y$ vs. $x$, find the rule relating $y$ and $x$.
Antidifferentiate: $y = \displaystyle\int 10e^{2x}\, dx = 10\cdot\tfrac{1}{2}e^{2x} + c = 5e^{2x} + c$
Apply the point $(0, 15)$: $15 = 5e^{0} + c = 5 + c$
$\therefore c = 10$
$y = 5e^{2x} + 10$
EXAMPLE 7
Given $\dfrac{dy}{dx} = \dfrac{2 - e^{5x}}{e^{3x}}$ and the graph of $y$ vs. $x$ passes through the origin, find the rule for $y$.
Split the fraction (you can't antidifferentiate a quotient directly):
$\dfrac{dy}{dx} = \dfrac{2}{e^{3x}} - \dfrac{e^{5x}}{e^{3x}} = 2e^{-3x} - e^{2x}$
$\dfrac{dy}{dx} = \dfrac{2}{e^{3x}} - \dfrac{e^{5x}}{e^{3x}} = 2e^{-3x} - e^{2x}$
Antidifferentiate each term:
$y = 2 \cdot \dfrac{1}{-3}e^{-3x} - \dfrac{1}{2}e^{2x} + c = -\dfrac{2}{3}e^{-3x} - \dfrac{1}{2}e^{2x} + c$
$y = 2 \cdot \dfrac{1}{-3}e^{-3x} - \dfrac{1}{2}e^{2x} + c = -\dfrac{2}{3}e^{-3x} - \dfrac{1}{2}e^{2x} + c$
Apply $(0, 0)$: $0 = -\dfrac{2}{3}e^{0} - \dfrac{1}{2}e^{0} + c = -\dfrac{2}{3} - \dfrac{1}{2} + c$
$c = \dfrac{2}{3} + \dfrac{1}{2} = \dfrac{4}{6} + \dfrac{3}{6} = \dfrac{7}{6}$
$y = -\dfrac{2}{3}e^{-3x} - \dfrac{1}{2}e^{2x} + \dfrac{7}{6}$
📺 Example 8 walkthrough: The two-step process — use the stationary point to find $k$ first, then antidifferentiate and use the point to find $c$.
EXAMPLE 8
Given $f'(x) = e^{2x} + k$ and $f(x)$ has a stationary point at $(0, 2)$, find $f(1)$.
Step 1 — use the stationary point to find $k$.
A stationary point at $x = 0$ means $f'(0) = 0$.
$f'(0) = e^{0} + k = 1 + k = 0 \;\;\Rightarrow\;\; k = -1$
A stationary point at $x = 0$ means $f'(0) = 0$.
$f'(0) = e^{0} + k = 1 + k = 0 \;\;\Rightarrow\;\; k = -1$
So $f'(x) = e^{2x} - 1$.
Step 2 — antidifferentiate to get $f(x)$.
$f(x) = \displaystyle\int (e^{2x} - 1)\, dx = \tfrac{1}{2}e^{2x} - x + c$
$f(x) = \displaystyle\int (e^{2x} - 1)\, dx = \tfrac{1}{2}e^{2x} - x + c$
Step 3 — use the point $(0, 2)$ to find $c$.
$2 = \tfrac{1}{2}e^{0} - 0 + c = \tfrac{1}{2} + c \;\;\Rightarrow\;\; c = \tfrac{3}{2}$
$2 = \tfrac{1}{2}e^{0} - 0 + c = \tfrac{1}{2} + c \;\;\Rightarrow\;\; c = \tfrac{3}{2}$
$\therefore f(x) = \tfrac{1}{2}e^{2x} - x + \tfrac{3}{2}$
Step 4 — evaluate $f(1)$.
$f(1) = \tfrac{1}{2}e^{2} - 1 + \tfrac{3}{2} = \tfrac{1}{2}e^{2} + \tfrac{1}{2}$
$f(1) = \tfrac{1}{2}e^{2} - 1 + \tfrac{3}{2} = \tfrac{1}{2}e^{2} + \tfrac{1}{2}$
$f(1) = \tfrac{1}{2}(e^{2} + 1) \;\;\approx\; 4.19$
Key result — trigonometric
FORMULAS
$\displaystyle\int \sin(kx+a)\, dx = -\tfrac{1}{k}\cos(kx+a) + c$
$\displaystyle\int \cos(kx+a)\, dx = \;\;\,\tfrac{1}{k}\sin(kx+a) + c$
$\displaystyle\int \cos(kx+a)\, dx = \;\;\,\tfrac{1}{k}\sin(kx+a) + c$
⚠️ Watch the sign: antidifferentiating $\sin$ gives a negative $\cos$ (and antidifferentiating $\cos$ gives a positive $\sin$). It's the opposite of differentiating $\cos \to -\sin$.
📺 Example 9 walkthrough: Trig antiderivative with initial condition — and the bonus identity $\cos(2x+\pi) = -\cos(2x)$ that simplifies the answer.
EXAMPLE 9
The function $f$ is such that $f'(x) = \sin(2x + \pi)$. If $f(0) = 4$, find the rule for $f$.
Use $\int \sin(kx+a)\, dx = -\tfrac{1}{k}\cos(kx+a) + c$ with $k=2$, $a=\pi$:
$f(x) = -\tfrac{1}{2}\cos(2x + \pi) + c$
Apply $f(0) = 4$: $4 = -\tfrac{1}{2}\cos(\pi) + c$
$\cos(\pi) = -1$, so $4 = -\tfrac{1}{2}(-1) + c = \tfrac{1}{2} + c$
$\therefore c = \tfrac{7}{2}$
$f(x) = -\tfrac{1}{2}\cos(2x + \pi) + \tfrac{7}{2}$
Bonus simplification: since $\cos(2x+\pi) = -\cos(2x)$, the rule can also be written $f(x) = \tfrac{1}{2}\cos(2x) + \tfrac{7}{2}$.
SIMILAR EXAMPLE 9
The graph of $f$ passes through $(0, 1)$, and $f'(x) = -\tfrac{3}{4}\sin(3x)$. Find the rule for $f$.
$f(x) = -\tfrac{3}{4}\displaystyle\int \sin(3x)\, dx$
$= -\tfrac{3}{4} \cdot \left(-\tfrac{1}{3}\right)\cos(3x) + c$
$= \tfrac{1}{4}\cos(3x) + c$
Apply $(0, 1)$: $1 = \tfrac{1}{4}\cos(0) + c = \tfrac{1}{4} + c$
$\therefore c = \tfrac{3}{4}$
$f(x) = \tfrac{1}{4}\cos(3x) + \tfrac{3}{4}$
Method summary
Initial condition problems — the recipe:
- Simplify first if needed (expand brackets, split quotients of $e^{kx}$).
- Antidifferentiate term by term — don't forget $+c$.
- Sub in the given point to solve for $c$.
- If there's an unknown constant inside $f'$ and a stationary point given, set $f'(\text{x-coord}) = 0$ first.
- Write the final rule clearly.
Working program — Cambridge textbook
After working through the examples above, complete in your exercise book:
| Exercise | Set work |
|---|---|
| 11B — The antiderivative | Q2c, Q3cf, Q4c, Half of Q5, Q6 |
| 11C (if time) | Q1 cfil, Q2–4 cf, Q5b, Q7c, Q9 |
Exit ticket — write in your book
Before you pack up, in your exercise book:
- State $\displaystyle\int \cos(kx+a)\, dx$.
- Find $\displaystyle\int (2e^x + 1)^2\, dx$ (hint: expand first).
- Given $f'(x) = \cos(2x)$ and $f\!\left(\tfrac{\pi}{4}\right) = 0$, find $f(x)$.