Area Under a Curve (Ex 11E / 11F / 11G)
Year 12 Mathematical Methods Unit 3 · Cambridge §11E/F/G · Mr Wong
Learning intentions.
- Evaluate definite integrals using the Fundamental Theorem of Calculus
- Use the properties of definite integrals (sum, swap, constant multiple)
- Distinguish signed area from total (geometric) area; split at every $x$-intercept
Key results
$\displaystyle\int_a^b f(x)\,dx = \bigl[G(x)\bigr]_a^b = G(b) - G(a)$ where $G$ is any antiderivative of $f$.
$\displaystyle\int_a^b k f \,dx = k\!\int_a^b f\,dx$ · $\displaystyle\int_a^b [f\pm g]\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$ · $\displaystyle\int_a^b f\,dx = -\!\int_b^a f\,dx$ · $\displaystyle\int_a^a f\,dx = 0$
For area below the axis, take an absolute value. For an interval crossing the axis, split at every $x$-intercept.
Warm-up — straight evaluations
1. Evaluate each definite integral. Show the bracket step.
a $\displaystyle\int_2^3 x^2\,dx$
b $\displaystyle\int_0^4 (e^{2x}+1)\,dx$
c $\displaystyle\int_6^8 \dfrac{1}{x-5}\,dx$
d $\displaystyle\int_0^{\pi/4} \sin(2x)\,dx$
Part A — Properties of the definite integral
2. Given $\displaystyle\int_1^4 f(x)\,dx = 3$, evaluate $\displaystyle\int_1^4 \bigl[2f(x) + 4\bigr]\,dx$.
3. If $\displaystyle\int_{-1}^{2} g(x)\,dx = 5$, find $\displaystyle\int_{-1}^{2} \bigl[x + 2g(x)\bigr]\,dx$.
4. If $\displaystyle\int_{-2}^{1} f(x)\,dx = 2$ and $\displaystyle\int_{1}^{3} f(x)\,dx = -6$, find $\displaystyle\int_{3}^{-2} f(x)\,dx$.
5. Find $k$ if $\displaystyle\int_0^k 8x\,dx = 36$ ($k>0$).
Area Under a Curve — continued
Part B (signed vs total area) · VCAA-style application
Part B — Signed area vs total area
6. Find $\displaystyle\int_4^9 \bigl(\sqrt{x}+1\bigr)\,dx$ and interpret as an area.
7. Calculate the area bounded by $y = x^2 - 4x$, the $x$-axis and the lines $x=1$, $x=3$.
(Hint: where is the curve relative to the axis on $[1,3]$?)
8. Let $f(x) = x^2 - 1$ for $x \in [0,2]$.
a Find the area bounded by $f$, the $x$-axis, the $y$-axis and $x=2$, by hand.
b A student types $\displaystyle\int_0^2 (x^2-1)\,dx$ into CAS and gets $\tfrac{2}{3}$. Explain why this is the wrong answer to the question in (a).
Part C — VCAA-style
9. VCAA 2016 Exam 1 (Tech-Free). Part of the graph of $f(x) = -x^2 + ax + 12$ is shown. The shaded region runs from $x=0$ to $x=3$ and has area $45$ square units. Find $a$, and then $m$ and $n$, the $x$-axis intercepts of $f$. (5 marks)
10. Find the total area bounded by $y = x^2 - 9$ and the $x$-axis between $x=0$ and $x=5$. (Hint: $x$-intercept inside the interval.)