Year 12 Methods — Integration Mixed Practice (11E

Today's lesson — pull it all together

This week you've covered the whole integration story: definite integrals (11E/F), areas under curves (11G), areas between curves (11I), integration by recognition (11H), and average value (11J). Today is a mixed-question session with all five topics jumbled together — exactly how the exam asks them.

Learning intentions

Identify which integration tool the question is asking for
Move fluently between area, signed integral, average value, and recognition
Set out solutions in the standard tech-free format expected on Exam 1

How to read an integration question — flowchart

DECISION TREE

Does the question say "find $\int \ldots dx$" with no diagram? → Evaluate the integral (signed). Use FTC.
Does it say "find the area" bounded by a curve and the $x$-axis? → Split at every $x$-intercept inside the interval; take absolute values on below-axis pieces.
Does it say "area enclosed by two curves"? → Find intersections; integrate top − bottom; split if upper curve swaps.
Does it give you a function and say "hence find $\int\ldots dx$"? → Recognition: differentiate, then rearrange.
Does it ask for the "average value" of $f$ on $[a,b]$? → $\bar y = \tfrac{1}{b-a}\int_a^b f(x)\,dx$. Watch for the shortcut on sinusoids over a full period.

Recap walkthrough — definite integral as signed area

📺 Walkthrough: $y = x^2 - 2x$ on $[0, 3]$ — signed integral is $0$ (red below the axis cancels green above), but the area is $\tfrac{4}{3} + \tfrac{2}{3} = 2$.

Worked examples — mixed

EXAMPLE A — total area

Find the total area bounded by $y = x^3 - 4x$ and the $x$-axis between $x = -2$ and $x = 2$.

$y = x(x^2 - 4) = x(x-2)(x+2)$ has roots at $x = -2, 0, 2$.

Test $x = -1$: $-1 + 4 = 3 > 0$ — positive on $(-2, 0)$. Test $x = 1$: $1 - 4 = -3 < 0$ — negative on $(0, 2)$.

By odd symmetry, $\bigl|\int_{-2}^0 y\,dx\bigr| = \bigl|\int_0^2 y\,dx\bigr|$. Compute one piece:

$\displaystyle\int_0^2 (x^3 - 4x)\,dx = \left[\tfrac{x^4}{4} - 2x^2\right]_0^2 = 4 - 8 = -4$

So each piece has area $4$. Total area $= 4 + 4$

$= 8$ square units

EXAMPLE B — area between curves

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

Intersect: $4 - x^2 = x + 2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0$. So $x = -2$ and $x = 1$.

Test $x = 0$: parabola gives $4$, line gives $2$. Parabola on top.

$A = \displaystyle\int_{-2}^{1} \bigl[(4 - x^2) - (x + 2)\bigr]\,dx = \int_{-2}^{1} (2 - x - x^2)\,dx$

$= \left[2x - \tfrac{x^2}{2} - \tfrac{x^3}{3}\right]_{-2}^{1} = \left(2 - \tfrac{1}{2} - \tfrac{1}{3}\right) - \left(-4 - 2 + \tfrac{8}{3}\right)$

$= \tfrac{7}{6} - \left(-\tfrac{10}{3}\right) = \tfrac{7}{6} + \tfrac{20}{6}$

$A = \dfrac{27}{6} = \dfrac{9}{2}$ square units

EXAMPLE C — integration by recognition

Let $f(x) = \log_e(x^2 + 1)$. (a) Find $f'(x)$. (b) Hence find $\displaystyle\int_0^1 \dfrac{x}{x^2 + 1}\,dx$ exactly.

(a) $f'(x) = \dfrac{2x}{x^2 + 1}$.

(b) $\displaystyle\int \dfrac{2x}{x^2+1}\,dx = \log_e(x^2 + 1) + c$. Divide by 2:

$\displaystyle\int_0^1 \dfrac{x}{x^2 + 1}\,dx = \tfrac{1}{2}\bigl[\log_e(x^2+1)\bigr]_0^1 = \tfrac{1}{2}(\log_e 2 - \log_e 1)$

$= \dfrac{1}{2}\log_e 2$

Recap walkthrough — average value

📺 Walkthrough: $f(x) = x^2$ on $[0, 3]$ — the gold rectangle at height $\bar y = 3$ has the same area as the region under the curve.

EXAMPLE D — average velocity (kinematics)

A particle's velocity (m/s) at time $t$ seconds is $v(t) = 3t^2$ for $0 \le t \le 4$. Find: (a) the displacement of the particle over the interval, and (b) the average velocity over the same interval.

(a) Displacement $= \displaystyle\int_0^4 v(t)\,dt = \int_0^4 3t^2\,dt = [t^3]_0^4 = 64$ m.

(b) Average velocity $= \bar v = \dfrac{1}{4 - 0}\displaystyle\int_0^4 v(t)\,dt = \dfrac{64}{4}$

$\bar v = 16$ m/s

Sanity check: at $t=0$, $v=0$; at $t=4$, $v=48$. Average velocity is weighted toward the larger speeds at the end of the interval — $16$ feels right (much more than $\tfrac{0+48}{2} = 24$? actually less — that's because $v$ grows fast, the time spent at small velocities is significant).

EXAMPLE E — average value of a sinusoid

Find the average value of $f(x) = 7\sin\!\left(\tfrac{x}{2}\right) - 3$ over the interval $[0, 4\pi]$.

Period of $\sin\!\left(\tfrac{x}{2}\right)$ is $\dfrac{2\pi}{1/2} = 4\pi$ — exactly one full period.

Over one full period the sine bit averages to $0$. So only the constant $-3$ remains.

$\bar y = -3$

Practice — pick the right tool

Practice P1 — area or signed integral?

(a) Evaluate $\displaystyle\int_{-1}^{2} (x^2 - 1)\,dx$. (b) Find the total area bounded by $y = x^2 - 1$, the $x$-axis, $x = -1$ and $x = 2$.

(a) $\left[\tfrac{x^3}{3} - x\right]_{-1}^{2} = \left(\tfrac{8}{3} - 2\right) - \left(-\tfrac{1}{3} + 1\right) = \tfrac{2}{3} - \tfrac{2}{3} = 0$.
(b) Roots at $x = \pm 1$. Inside the interval, only $x = 1$ is a root (and $x = -1$ is an endpoint). Below-axis on $[-1, 1]$: $\bigl|\int_{-1}^1 (x^2-1)\,dx\bigr| = \bigl|-\tfrac{4}{3}\bigr| = \tfrac{4}{3}$. Above-axis on $[1, 2]$: $\int_1^2 (x^2-1)\,dx = \tfrac{4}{3}$. Total area $= \dfrac{8}{3}$.

Practice P2 — recognition.

Let $f(x) = \sin^2(x)$. Find $f'(x)$, hence find $\displaystyle\int_0^{\pi/2} \sin x \cos x\,dx$.

$f'(x) = 2\sin x \cos x$. Hence $\int 2\sin x\cos x\,dx = \sin^2 x + c$, so $\int \sin x \cos x\,dx = \tfrac{1}{2}\sin^2 x + c$.
$\int_0^{\pi/2} \sin x\cos x\,dx = \tfrac{1}{2}\bigl[\sin^2 x\bigr]_0^{\pi/2} = \tfrac{1}{2}(1 - 0) = \dfrac{1}{2}$.

Practice P3 — average value.

The velocity of a car (m/s) is $v(t) = 20 - 2t$ for $0 \le t \le 10$. Find (a) the distance travelled, and (b) the average velocity.

(a) $v \ge 0$ on the whole interval (since $v(10) = 0$). Distance $= \int_0^{10} (20 - 2t)\,dt = [20t - t^2]_0^{10} = 200 - 100 = 100$ m.
(b) $\bar v = \dfrac{100}{10} = 10$ m/s. Equivalently: $v$ is linear from $20$ to $0$, average $= \tfrac{20+0}{2} = 10$ ✓.

Practice P4 — area between curves.

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

Intersect: $x^2 = 4x - x^2 \Rightarrow 2x^2 = 4x \Rightarrow x = 0, 2$. Test $x = 1$: top curve $= 4 - 1 = 3$, bottom $= 1$. The "upside-down" parabola $4x - x^2$ is on top.
$A = \int_0^2 \bigl[(4x - x^2) - x^2\bigr]\,dx = \int_0^2 (4x - 2x^2)\,dx = \bigl[2x^2 - \tfrac{2x^3}{3}\bigr]_0^2 = 8 - \tfrac{16}{3} = \dfrac{8}{3}$ sq units.

Quick quiz (5 min) — mixed

Working program — Cambridge Ch 11 review

After the quiz, open Chapter 11 review pages:

Section	Set work
Chapter 11 review — Technology-free	Multiple-choice Q1–6, Short-answer Q1, Q3, Q5
Chapter 11 review — Technology-active	Extended-response Q1, Q2 (areas, average value applications)
Past VCAA Exam 1 questions	2007 Q9, 2013 (recognition), 2016 Q? (area)

This wraps the Integration chapter. Next week we move to the next chapter — bring your CAS and a clean exercise book.

Exit ticket — write in your book

Before you pack up, in your exercise book:

Write the five integration "tools" you've learned this chapter, one sentence each.
One question that's still bothering you — bring it to next lesson.
Self-rate (1 → 5) your confidence on (a) areas, (b) recognition, (c) average value. Honest answers, please.