Find antiderivatives, evaluate definite integrals using the Fundamental Theorem of Calculus, and apply integration to areas, total change and kinematics

What this dot point is asking

SCSA Unit 4 builds the integral calculus that complements the differentiation of Unit 3. You learn to antidifferentiate standard functions, evaluate definite integrals, and connect the definite integral to signed area and to the accumulation of a rate of change. This dot point appears in both the calculator-free and calculator-assumed sections of the WACE written examination.

Antiderivatives

Antidifferentiation reverses the chain rule for linear inner functions: the factor $\frac{1}{k}$ appears because differentiating $e^{kx}$ or $\sin(kx)$ produces an extra $k$. The constant of integration $c$ must be written on every indefinite integral.

The definite integral and the Fundamental Theorem of Calculus

The definite integral $\int_a^b f(x)\,dx$ gives the signed area between the curve and the $x$-axis from $x = a$ to $x = b$: regions above the axis count as positive and regions below as negative. Because the constant $c$ appears in both $F(b)$ and $F(a)$ it cancels on subtraction, which is why a definite integral has a single numerical value while an indefinite integral is a whole family of curves. This cancellation is the practical reason the constant of integration is never written when evaluating a definite integral.

Area under and between curves

The area enclosed between two curves $y = f(x)$ (upper) and $y = g(x)$ (lower) over $[a, b]$ is

where $a$ and $b$ are usually the $x$-coordinates of the points of intersection. Always subtract the lower function from the upper function so the integrand is positive.

Total change and kinematics

If $f(x)$ is a rate of change, then $\int_a^b f(x)\,dx$ is the net change in the quantity over $[a, b]$. In kinematics, with velocity $v(t)$ and acceleration $a(t)$:

• $\displaystyle\int v(t)\,dt = x(t) + c$ gives displacement from velocity.
• $\displaystyle\int a(t)\,dt = v(t) + c$ gives velocity from acceleration.
• Displacement over $[t_1, t_2]$ is $\displaystyle\int_{t_1}^{t_2} v(t)\,dt$ (signed), while distance travelled is $\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt$.

For example, a particle with velocity $v(t) = 4 - 2t$ (in m/s) has displacement over $0 \le t \le 3$ equal to $\int_0^3 (4 - 2t)\,dt = [4t - t^2]_0^3 = 12 - 9 = 3$ m. To find distance travelled you would split the integral at $t = 2$ where $v = 0$, because the particle reverses direction.