Today's lesson
You can already antidifferentiate (Chapter 11B). Today we turn that into a number: the definite integral, which gives the exact area under a curve.
Learning intentions
- Evaluate a definite integral using the Fundamental Theorem of Calculus
- Use the properties of definite integrals (sum, swap limits, constant multiple)
- Recognise that the integral gives the signed area — regions below the $x$-axis contribute a negative amount
- Compute the total (geometric) area of a region that crosses the $x$-axis by splitting at every $x$-intercept
Part 1 — The Fundamental Theorem of Calculus
FUNDAMENTAL THEOREM OF CALCULUS
If $f$ is continuous on $[a,b]$ and $G$ is any antiderivative of $f$, then
$$\int_a^b f(x)\,dx \;=\; \Bigl[G(x)\Bigr]_a^b \;=\; G(b) - G(a).$$
The square-bracket notation $\bigl[G(x)\bigr]_a^b$ just means "evaluate at the top minus evaluate at the bottom". The $+c$ cancels when you subtract, so leave it out of definite integrals.
📺 Walkthrough: evaluating $\displaystyle\int_1^3 \left(\tfrac{x^2}{2}+1\right) dx$ as the shaded area, working from the antiderivative to the answer $\tfrac{22}{3}$.
Part 2 — Warm-up evaluations
Use the antiderivative tables you already know from §11B.
WARM-UP (a)
Evaluate $\displaystyle\int_2^3 x^2\, dx$.
$=\left[\tfrac{x^3}{3}\right]_2^3$
$=\tfrac{27}{3} - \tfrac{8}{3}$
$=\dfrac{19}{3}$
WARM-UP (b)
Evaluate $\displaystyle\int_0^4 \bigl(e^{2x}+1\bigr) dx$.
$=\left[\tfrac{1}{2}e^{2x} + x\right]_0^4$
$=\left(\tfrac{1}{2}e^8 + 4\right) - \left(\tfrac{1}{2}e^0 + 0\right)$
$=\tfrac{1}{2}e^8 + 4 - \tfrac{1}{2}$
$=\dfrac{1}{2}e^{8} + \dfrac{7}{2}$
WARM-UP (c)
Evaluate $\displaystyle\int_6^8 \dfrac{1}{x-5}\, dx$.
$=\bigl[\log_e(x-5)\bigr]_6^8$ (since $x-5>0$)
$=\log_e 3 - \log_e 1$
$=\log_e 3$
WARM-UP (d)
Evaluate $\displaystyle\int_0^{\pi/4} \sin(2x)\, dx$.
$=\left[-\tfrac{1}{2}\cos(2x)\right]_0^{\pi/4}$
$=-\tfrac{1}{2}\cos\!\left(\tfrac{\pi}{2}\right) - \!\left(-\tfrac{1}{2}\cos 0\right)$
$=-\tfrac{1}{2}(0) + \tfrac{1}{2}(1)$
$=\dfrac{1}{2}$
Part 3 — Properties of the definite integral
Five useful properties
- Split the interval: $\displaystyle\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx$
- Same endpoints: $\displaystyle\int_a^a f\,dx = 0$
- Constant multiple: $\displaystyle\int_a^b k f(x)\,dx = k\int_a^b f(x)\,dx$
- Sum/difference: $\displaystyle\int_a^b \bigl[f \pm g\bigr]\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$
- Swap limits — change sign: $\displaystyle\int_a^b f\,dx = -\int_b^a f\,dx$
EXAMPLE 5(a) — using properties
Given $\displaystyle\int_1^4 f(x)\, dx = 3$, find $\displaystyle\int_1^4 \bigl[2f(x) + 4\bigr] dx$ without knowing what $f$ is.
Split using sum and constant-multiple properties:
$= 2\displaystyle\int_1^4 f(x)\, dx + \int_1^4 4\, dx$
$= 2(3) + 4(4-1)$ (the $\int 4\,dx$ over an interval of length $3$ is $12$)
$= 6 + 12$
$= 18$
SIMILAR EXAMPLE 6
If $\displaystyle\int_{-1}^{2} g(x)\, dx = 5$, find $\displaystyle\int_{-1}^{2} \bigl[x + 2g(x)\bigr] dx$.
Split: $=\displaystyle\int_{-1}^{2} x\, dx + 2\!\int_{-1}^{2} g(x)\, dx$
$=\left[\tfrac{x^2}{2}\right]_{-1}^{2} + 2(5)$
$=\left(\tfrac{4}{2} - \tfrac{1}{2}\right) + 10 = \tfrac{3}{2} + 10$
$=\dfrac{23}{2}$
SIMILAR EXAMPLE 7
If $\displaystyle\int_{-2}^{1} f(x)\, dx = 2$ and $\displaystyle\int_{1}^{3} f(x)\, dx = -6$, find $\displaystyle\int_{3}^{-2} f(x)\, dx$.
Join the two known pieces: $\displaystyle\int_{-2}^{3} f = \int_{-2}^{1} f + \int_{1}^{3} f = 2 + (-6) = -4$
Now swap the limits (changes the sign):
$\displaystyle\int_{3}^{-2} f = -\int_{-2}^{3} f = -(-4)$
$= 4$
Part 4 — Signed area vs total area
The definite integral gives the signed area — anything below the $x$-axis counts as negative. For the geometric (total) area you must take absolute values of any below-axis pieces.
📺 Walkthrough: $y=\sin x$ over $[-\pi,\pi]$. Signed area is $0$ (the bits cancel); total area is $4$ (flip the negative bit up).
⚠️ "Find the area" vs "evaluate the integral". If the question says find the area (or talks about an enclosed region), you must give a positive number — split at every $x$-intercept inside the interval and add absolute values.
TOTAL AREA RECIPE
For a region bounded by $y=f(x)$, the $x$-axis and the lines $x=a$, $x=b$:
- Find every $x$-intercept of $f$ that lies in $(a,b)$ — call them $c_1, c_2, \ldots$
- On each sub-interval, take $\left|\displaystyle\int f\, dx\right|$
- Total area = sum of those absolute values
EXAMPLE 1 — area above the axis
Find $\displaystyle\int_4^9 \bigl(\sqrt{x}+1\bigr) dx$ and illustrate the result as an area under the graph.
$\sqrt{x} = x^{1/2}$, so antiderivative is $\tfrac{2}{3}x^{3/2} + x$.
$=\left[\tfrac{2}{3}x^{3/2} + x\right]_4^9$
$=\left(\tfrac{2}{3}\cdot 27 + 9\right) - \left(\tfrac{2}{3}\cdot 8 + 4\right)$
$=(18+9) - \!\left(\tfrac{16}{3}+4\right) = 27 - \tfrac{28}{3} = \tfrac{81-28}{3}$
$=\dfrac{53}{3}$ (the entire curve is above the axis on $[4,9]$, so signed area = total area).
EXAMPLE 2 — area below the axis
Calculate the shaded area bounded by $y = x^2 - 4x$, the $x$-axis and the lines $x=1$ and $x=3$.
Check sign first. $y = x(x-4)$ has roots at $0$ and $4$. Between $x=1$ and $x=3$ the curve is below the $x$-axis (try $x=2$: $y = 4 - 8 = -4$).
So the signed integral will be negative. Take an absolute value at the end.
$\displaystyle\int_1^3 (x^2 - 4x)\, dx = \left[\tfrac{x^3}{3} - 2x^2\right]_1^3$
$=\left(9 - 18\right) - \left(\tfrac{1}{3} - 2\right) = -9 - \tfrac{1}{3} + 2 = -\tfrac{22}{3}$
Area (positive) $= \left|-\tfrac{22}{3}\right|$
Area $= \dfrac{22}{3}$ square units
EXAMPLE 3 — region crossing the axis
Let $f(x) = x^2 - 1$ for $x \in [0,2]$.
(a) Find the area bounded by the graph of $f$, the $x$-axis, the $y$-axis and the line $x=2$, by hand.
(b) Explain why $\displaystyle\int_0^2 (x^2-1)\,dx = \tfrac{2}{3}$ on CAS is the wrong answer for this area.
(a) $f$ has an $x$-intercept at $x=1$. From $0$ to $1$, $f<0$; from $1$ to $2$, $f>0$. Split there.
Below-axis piece (take absolute value):
$\left|\displaystyle\int_0^1 (x^2-1)\,dx\right| = \left|\left[\tfrac{x^3}{3} - x\right]_0^1\right| = \left|\tfrac{1}{3} - 1\right| = \tfrac{2}{3}$
Above-axis piece:
$\displaystyle\int_1^2 (x^2-1)\,dx = \left[\tfrac{x^3}{3} - x\right]_1^2 = \!\left(\tfrac{8}{3}-2\right) - \!\left(\tfrac{1}{3}-1\right) = \tfrac{2}{3} + \tfrac{2}{3} = \tfrac{4}{3}$
Total area $= \tfrac{2}{3} + \tfrac{4}{3}$
Area $= 2$ square units
(b) CAS just evaluated $\displaystyle\int_0^2 (x^2-1)\,dx = \tfrac{4}{3} - \tfrac{2}{3} = \tfrac{2}{3}$ — the signed area. The negative bit from $[0,1]$ has cancelled most of the positive bit from $[1,2]$. For a geometric area you must split at $x=1$ and add absolute values, as in part (a).
Practice 4.1 — definite integrals (tech-free).
- $\displaystyle\int_0^2 (3x^2 + 2)\,dx$
- $\displaystyle\int_0^{\pi} \cos x\,dx$
- $\displaystyle\int_{1}^{e} \dfrac{1}{x}\,dx$
- $\displaystyle\int_0^1 e^{2x}\,dx$
a) $\left[x^3 + 2x\right]_0^2 = 8 + 4 = 12$
b) $\left[\sin x\right]_0^\pi = \sin\pi - \sin 0 = 0 - 0 = 0$ (the positive bump on $[0,\pi/2]$ cancels the negative one on $[\pi/2,\pi]$).
c) $\left[\log_e x\right]_1^e = 1 - 0 = 1$
d) $\left[\tfrac{1}{2}e^{2x}\right]_0^1 = \tfrac{1}{2}(e^2 - 1)$
b) $\left[\sin x\right]_0^\pi = \sin\pi - \sin 0 = 0 - 0 = 0$ (the positive bump on $[0,\pi/2]$ cancels the negative one on $[\pi/2,\pi]$).
c) $\left[\log_e x\right]_1^e = 1 - 0 = 1$
d) $\left[\tfrac{1}{2}e^{2x}\right]_0^1 = \tfrac{1}{2}(e^2 - 1)$
Practice 4.2 — area, not signed integral.
Find the total area bounded by $y = x^2 - 9$ and the $x$-axis between $x = 0$ and $x = 5$.
$x$-intercept at $x = 3$ (in the interval). Split at $3$.
$\displaystyle\int_0^3 (x^2-9)\,dx = \left[\tfrac{x^3}{3} - 9x\right]_0^3 = 9 - 27 = -18$ → take absolute value: $18$
$\displaystyle\int_3^5 (x^2-9)\,dx = \left[\tfrac{x^3}{3} - 9x\right]_3^5 = \!\left(\tfrac{125}{3}-45\right) - (-18) = -\tfrac{10}{3} + 18 = \tfrac{44}{3}$
Total area $= 18 + \tfrac{44}{3} = \dfrac{98}{3}$ square units.
$\displaystyle\int_0^3 (x^2-9)\,dx = \left[\tfrac{x^3}{3} - 9x\right]_0^3 = 9 - 27 = -18$ → take absolute value: $18$
$\displaystyle\int_3^5 (x^2-9)\,dx = \left[\tfrac{x^3}{3} - 9x\right]_3^5 = \!\left(\tfrac{125}{3}-45\right) - (-18) = -\tfrac{10}{3} + 18 = \tfrac{44}{3}$
Total area $= 18 + \tfrac{44}{3} = \dfrac{98}{3}$ square units.
Part 5 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. $\displaystyle\int_0^2 (4x-3)\,dx \;=$
Q2. If $\displaystyle\int_1^5 f(x)\,dx = 8$, then $\displaystyle\int_5^1 f(x)\,dx \;=$
Q3. $\displaystyle\int_0^{\pi/2} \sin x\,dx \;=$
Q4. Which equals the area bounded by $y = x^2 - 4$, the $x$-axis between $x=0$ and $x=2$?
Q5. Given $\displaystyle\int_0^3 f = 5$ and $\displaystyle\int_0^3 g = 2$, $\displaystyle\int_0^3 \bigl[3f - 2g\bigr] dx \;=$
Working program — Cambridge §11E, 11F, 11G
After the quiz, open Chapter 11 and work through these:
| Exercise | Set work |
|---|---|
| 11E — The definite integral | Q1 acef, Q2 ace, Q3, Q5 |
| 11F — Properties of definite integrals | Q1, Q3, Q5, Q7 |
| 11G — Area under a curve | Q1 ace, Q2, Q3, Q5 |
If you finish early, start 11I — Area between curves (today's second lesson).
Exit ticket — write in your book
Before you pack up, in your exercise book:
- State the Fundamental Theorem of Calculus in your own words (one sentence).
- Evaluate $\displaystyle\int_0^1 (3x^2 + 2x)\,dx$.
- True or false: "$\displaystyle\int_a^b f(x)\,dx$ is always the area between the graph and the $x$-axis." Explain.