Calculus (SL) — IB Math AA

1. Limits & Continuity

A limit describes the value a function approaches as the input approaches a particular point:

$$\lim_{x o a} f(x) = L$$

means $f(x)$ gets arbitrarily close to $L$ as $x$ approaches $a$ (but need not equal $L$ at $x = a$).

A function $f$ is continuous at $x = a$ if three conditions hold:

$f(a)$ is defined
$\lim_{x o a} f(x)$ exists
$\lim_{x o a} f(x) = f(a)$

Practical limits: For most functions (polynomials, rationals, trig, exponential), simply substitute $x = a$. If you get $\frac{0}{0}$, factor and cancel, or use L'Hôpital's rule (HL).

2. Differentiation Rules

The derivative $f'(x)$ gives the instantaneous rate of change of $f$ at $x$, and equals the gradient of the tangent to $y = f(x)$ at that point.

Standard Derivatives

$\dfrac{d}{dx}[x^n] = nx^{n-1}$

$\dfrac{d}{dx}[e^x] = e^x$

$\dfrac{d}{dx}[\ln x] = \dfrac{1}{x}$

$\dfrac{d}{dx}[\sin x] = \cos x$

$\dfrac{d}{dx}[\cos x] = -\sin x$

$\dfrac{d}{dx}[\tan x] = \sec^2 x$

Differentiation Rules

Sum/Difference: $(f \pm g)' = f' \pm g'$
Product Rule: $(fg)' = f'g + fg'$
Quotient Rule: $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$
Chain Rule: $\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

3. Applications of Differentiation

Tangent and Normal Lines

At the point $(a, f(a))$:

Tangent line gradient: $m_T = f'(a)$, equation: $y - f(a) = f'(a)(x - a)$
Normal line gradient: $m_N = -\dfrac{1}{f'(a)}$ (perpendicular to tangent)

Increasing, Decreasing & Stationary Points

$f'(x) > 0$: $f$ is increasing on that interval
$f'(x) < 0$: $f$ is decreasing on that interval
$f'(x) = 0$: stationary point (could be local max, local min, or inflection)

First Derivative Test for Local Extrema

At a stationary point $x = c$ (where $f'(c) = 0$):

If $f'$ changes from $+$ to $-$: local maximum
If $f'$ changes from $-$ to $+$: local minimum
If $f'$ does not change sign: point of inflection

Points of Inflection & Second Derivative

A point of inflection is where concavity changes ($f''(x) = 0$ and changes sign).

Second derivative test: if $f'(c)=0$ and $f''(c) < 0$, then local max; if $f''(c) > 0$, then local min.

4. Integration

Integration is the reverse of differentiation. The indefinite integral includes an arbitrary constant $C$.

Standard Integrals

$\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)$

$\displaystyle\int e^x\,dx = e^x + C$

$\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C$

$\displaystyle\int \sin x\,dx = -\cos x + C$

$\displaystyle\int \cos x\,dx = \sin x + C$

Integration by Substitution

For $\displaystyle\int f(g(x)) \cdot g'(x)\,dx$, let $u = g(x)$, then $du = g'(x)\,dx$:

$$\int f(g(x)) \cdot g'(x)\,dx = \int f(u)\,du$$

5. Definite Integrals & Area

Fundamental Theorem of Calculus

$$\int_a^b f(x)\,dx = F(b) - F(a)$$

where $F$ is any antiderivative of $f$ (i.e., $F' = f$).

Area Between Curves

The area enclosed between $y = f(x)$ (upper) and $y = g(x)$ (lower), from $x = a$ to $x = b$:

$$A = \int_a^b [f(x) - g(x)]\,dx$$

If the curves intersect, find the intersection points first and split the integral as needed.

Sign of the integral: $\int_a^b f(x)\,dx$ can be negative if $f(x) < 0$ on $[a,b]$. For area, always take the absolute value of each section, or use $\int_a^b |f(x)|\,dx$.

6. Kinematics

Calculus connects the three kinematic quantities: displacement $s$, velocity $v$, and acceleration $a$.

Kinematic Relationships

$$v = \frac{ds}{dt} = s'(t) \qquad a = \frac{dv}{dt} = v'(t) = s''(t)$$

$$s = \int v\,dt \qquad v = \int a\,dt$$

Particle at rest: $v = 0$
Particle changes direction: $v$ changes sign (and equals 0 at that instant)
Total distance $\neq$ displacement when particle reverses direction

Worked Examples

Worked Example 1

Finding Local Maximum and Minimum Values

Find the local maxima and minima of $f(x) = 2x^3 - 9x^2 + 12x - 4$.

Differentiate: $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$

Set $f'(x) = 0$: $x = 1$ or $x = 2$ (stationary points).

Sign of $f'(x)$: positive for $x < 1$, negative for $1 < x < 2$, positive for $x > 2$.
At $x=1$: $f'$ changes $+ o -$ → local maximum. $f(1) = 2-9+12-4 = 1$.
At $x=2$: $f'$ changes $- o +$ → local minimum. $f(2) = 16-36+24-4 = 0$.

Worked Example 2

Definite Integral — Area Under a Curve

Find the area enclosed by $y = x^2 - 4$ and the $x$-axis.

Find where the curve crosses the $x$-axis: $x^2 - 4 = 0 \implies x = \pm 2$.

On $[-2, 2]$, $y = x^2 - 4 \leq 0$, so the area is given by $A = -\displaystyle\int_{-2}^{2}(x^2-4)\,dx$.

$\displaystyle\int_{-2}^{2}(x^2-4)\,dx = \left[\frac{x^3}{3} - 4x\right]_{-2}^{2} = \left(\frac{8}{3}-8\right) - \left(\frac{-8}{3}+8\right) = \frac{8}{3}-8+\frac{8}{3}-8 = \frac{16}{3} - 16 = -\frac{32}{3}$

Area $= -\!\left(-\dfrac{32}{3}\right) = \boxed{\dfrac{32}{3}}$ square units.

Worked Example 3

Kinematics — Displacement, Velocity, and Total Distance

A particle moves along a straight line with velocity $v(t) = t^2 - 5t + 6$ m/s, $t \geq 0$. (a) Find when the particle is at rest. (b) Find the total distance travelled in the first 4 seconds.

Part (a): At rest when $v = 0$: $t^2 - 5t + 6 = (t-2)(t-3) = 0 \implies t = 2$ s or $t = 3$ s.

Check signs: $v > 0$ on $[0,2)$, $v < 0$ on $(2,3)$, $v > 0$ on $(3,4]$. Particle reverses at $t=2$ and $t=3$.

Part (b): Total distance $= \displaystyle\int_0^2 v\,dt + \left|\int_2^3 v\,dt\right| + \int_3^4 v\,dt$

$\displaystyle\int v\,dt = \frac{t^3}{3} - \frac{5t^2}{2} + 6t + C$.
$\int_0^2 v\,dt = \left(\frac{8}{3}-10+12\right) - 0 = \frac{8}{3}+2 = \frac{14}{3}$
$\int_2^3 v\,dt = \left(9-\frac{45}{2}+18\right) - \left(\frac{8}{3}-10+12\right) = \frac{9}{2} - \frac{14}{3} = \frac{27-28}{6} = -\frac{1}{6}$
$\int_3^4 v\,dt = \left(\frac{64}{3}-40+24\right) - \frac{9}{2} = \frac{64}{3}-16-\frac{9}{2} = \frac{128-96-27}{6} = \frac{5}{6}$

Total distance $= \dfrac{14}{3} + \dfrac{1}{6} + \dfrac{5}{6} = \dfrac{28}{6} + \dfrac{1}{6} + \dfrac{5}{6} = \dfrac{34}{6} = \boxed{\dfrac{17}{3}}$ m $\approx 5.67$ m.

Practice Problems

Q1. Differentiate $f(x) = (3x^2 + 1)^4$ using the chain rule.

Show Solution

Let $u = 3x^2 + 1$, so $f = u^4$.

$\dfrac{du}{dx} = 6x$ and $\dfrac{df}{du} = 4u^3$.

$f'(x) = 4u^3 \cdot 6x = 4(3x^2+1)^3 \cdot 6x = \mathbf{24x(3x^2+1)^3}$

Q2. Find the equation of the tangent to $y = x^3 - 3x + 2$ at the point where $x = 2$.

Show Solution

At $x = 2$: $y = 8 - 6 + 2 = 4$. Point: $(2, 4)$.

$y' = 3x^2 - 3$. At $x = 2$: $m = 3(4)-3 = 9$.

Tangent: $y - 4 = 9(x - 2) \implies \mathbf{y = 9x - 14}$.

Q3. Evaluate $\displaystyle\int_0^{\pi/2} (2\sin x + 3\cos x)\,dx$.

Show Solution

$\displaystyle\int_0^{\pi/2}(2\sin x + 3\cos x)\,dx = \big[-2\cos x + 3\sin x\big]_0^{\pi/2}$

$= \big(-2\cos\frac{\pi}{2} + 3\sin\frac{\pi}{2}\big) - \big(-2\cos 0 + 3\sin 0\big)$

$= (0 + 3) - (-2 + 0) = 3 + 2 = \mathbf{5}$

Q4. The acceleration of a particle is $a(t) = 6t - 2$ m/s². The particle has velocity $v = 3$ m/s at $t = 0$. Find $v(t)$ and the displacement from $t = 0$ to $t = 3$.

Show Solution

$v(t) = \displaystyle\int a\,dt = 3t^2 - 2t + C$. Using $v(0) = 3$: $C = 3$. So $v(t) = 3t^2 - 2t + 3$.

Displacement $= \displaystyle\int_0^3 v\,dt = \big[t^3 - t^2 + 3t\big]_0^3 = (27 - 9 + 9) - 0 = \mathbf{27}$ m.