The Definite Integral and Area

For $f : (-\infty, 1] \to R$, $f(x) = x^2 - 2x$ and its inverse $f^{-1}$, calculate the area of the regions enclosed by the curves of $f$, $f^{-1}$ and $y = -x$.

1. 1

   By symmetry in $y = x$ the two enclosed regions are equal. Find where $f$ meets $y = -x$.

   $$x^2 - 2x = -x \implies x^2 - x = 0 \implies x = 0 \text{ or } x = 1$$
2. 2

   One region's area is the integral of top minus bottom between $0$ and $1$.

   $$\int_0^1 \big((-x) - (x^2 - 2x)\big)\,dx = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{6}$$
3. 3

   Double for the two identical regions. The report (a low scoring part) flags sign errors from dropping brackets, and students forgetting the second region.

   $$2 \times \frac{1}{6} = \frac{1}{3}$$

VCAA 2023 Mathematical Methods Exam 1, Q7d

The power, in kilowatts, used by a heater $t$ hours after it is switched on is $p(t) = 1.5$ for $0 \le t \le 0.4$ and $p(t) = 0.3 + Ae^{-10t}$ for $t > 0.4$. Energy (kilowatt hours) is the area between $y = p(t)$ and the $t$ axis. (i) Given $p(t)$ is continuous for $t \ge 0$, show that $A = 1.2e^4$. (ii) Find how long until the heater has used $0.5$ kilowatt hours of energy. (iii) Find how long until it has used $1$ kilowatt hour of energy, correct to two decimal places.

1. 1

   (i) Continuity at $t = 0.4$ means the two branches agree there.

   $$1.5 = 0.3 + Ae^{-10(0.4)} \implies Ae^{-4} = 1.2 \implies A = 1.2e^{4}$$
2. 2

   (ii) For $t \le 0.4$ the power is constant, so energy is $1.5t$. Set this to $0.5$, valid since $\tfrac13 < 0.4$.

   $$1.5t = 0.5 \implies t = \frac{1}{3} \text{ hours}$$
3. 3

   (iii) Energy up to $t = 0.4$ is $0.6$ kWh; integrate the second branch and solve for a total of $1$ kWh. Some students solved $p(t) = 0.5$ instead of using the area.

   $$0.6 + \int_{0.4}^{T}\big(0.3 + 1.2e^4 e^{-10t}\big)\,dt = 1 \implies T \approx 1.33 \text{ hours}$$

VCAA 2024 Mathematical Methods Exam 2, Section B Q2f

For $g(x) = \dfrac{1}{(x+3)^2} - 2$, determine the area of the region bounded by the line $x = -2$, the $x$ axis, the $y$ axis and the graph of $y = g(x)$.

1. 1

   The graph lies below the $x$ axis on $[-2, 0]$, so the area is the absolute value of the integral.

   $$A = \left|\int_{-2}^{0}\left(\frac{1}{(x+3)^2}-2\right)dx\right|$$
2. 2

   Antidifferentiate term by term.

   $$\int\left(\frac{1}{(x+3)^2}-2\right)dx = -\frac{1}{x+3}-2x$$
3. 3

   Evaluate between $-2$ and $0$, then take the magnitude. The report notes students wrongly used an $x$ intercept as a terminal, or left the answer as the negative $-\tfrac{10}{3}$.

   $$A = \left|\left(-\tfrac{1}{3}\right)-\left(3\right)\right| = \frac{10}{3}$$

VCAA 2024 Mathematical Methods Exam 1, Q3b

Part of the graph of $f : [-\pi, \pi] \to R,\ f(x) = x\sin(x)$ is shown. Use the trapezium rule with a step size of $\dfrac{\pi}{3}$ to determine an approximation of the total area between the graph of $y = f(x)$ and the $x$ axis over the interval $x \in [0, \pi]$.

1. 1

   Find the function values at $x = 0, \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \pi$.

   $$f(0)=0,\ f\!\left(\tfrac{\pi}{3}\right)=\frac{\pi\sqrt3}{6},\ f\!\left(\tfrac{2\pi}{3}\right)=\frac{\pi\sqrt3}{3},\ f(\pi)=0$$
2. 2

   Apply the trapezium rule with $h = \tfrac{\pi}{3}$. The report flags omitting the coefficient $2$ on the middle terms as the common error.

   $$A \approx \frac{\pi}{6}\left[0 + 2\cdot\frac{\pi\sqrt3}{6} + 2\cdot\frac{\pi\sqrt3}{3} + 0\right]$$
3. 3

   Simplify. Integral calculus was not acceptable here.

   $$A = \frac{\pi}{6}\left(\pi\sqrt3\right) = \frac{\sqrt3\,\pi^2}{6}$$

VCAA 2024 Mathematical Methods Exam 1, Q7a

Evaluate $\displaystyle\int_{1}^{3} (2x + 1)\,dx$.

1. 1

   Antidifferentiate each term. Raise the power by one and divide by the new power. No $+c$ is needed for a definite integral.

   $$\int (2x + 1)\,dx = x^2 + x$$
2. 2

   Write the antiderivative in square brackets with the terminals on the right.

   $$\Big[\, x^2 + x \,\Big]_{1}^{3}$$
3. 3

   Substitute the top terminal, then subtract the bottom terminal. Keep brackets around each piece.

   $$(3^2 + 3) - (1^2 + 1) = 12 - 2$$

Find the exact area bounded by $y = 3x^2$, the $x$ axis, and the lines $x = 0$ and $x = 2$.

1. 1

   The curve sits above the $x$ axis on this interval, so the area is just the definite integral.

   $$A = \int_{0}^{2} 3x^2 \,dx$$
2. 2

   Antidifferentiate, then evaluate at the terminals.

   $$\Big[\, x^3 \,\Big]_{0}^{2} = (2)^3 - (0)^3$$
3. 3

   The region is above the axis, so the answer is already positive. State the units of area if given.

   $$8 - 0 = 8$$

Find the area bounded by $y = x^2 - 4$, the $x$ axis, and the lines $x = 0$ and $x = 2$.

1. 1

   Check the sign first. On $0 \le x \le 2$ the curve lies below the $x$ axis, so the integral will be negative.

   $$\int_{0}^{2} (x^2 - 4)\,dx$$
2. 2

   Antidifferentiate and evaluate. This gives the signed value.

   $$\Big[\, \tfrac{1}{3}x^3 - 4x \,\Big]_{0}^{2} = \big(\tfrac{8}{3} - 8\big) - 0 = -\tfrac{16}{3}$$
3. 3

   Area is a positive quantity. Take the absolute value of the negative signed result.

   $$A = \left| -\tfrac{16}{3} \right| = \tfrac{16}{3}$$