A car's speed varies throughout a journey. The speedometer shows instantaneous speed, but the odometer shows total distance. The definite integral is the mathematical odometer: it adds up all the tiny pieces of distance travelled during each instant. By placing upper and lower limits on the integral, we turn the abstract antiderivative into a concrete, numerical answer.
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A car travels at $v(t) = 2t$ m/s. Without calculating, do you think the distance travelled in the first 3 seconds is more than, less than, or equal to $2 \times 3 = 6$ metres? Think about whether the car is speeding up or slowing down.
Core Content
A definite integral has upper and lower limits and produces a number (not a function):
โซ_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)
where $F(x)$ is any antiderivative of $f(x)$.
Why no +C? Suppose we used $F(x) + C$:
[F(x) + C]_a^b = (F(b) + C) - (F(a) + C) = F(b) - F(a)
The constants cancel! This is why definite integrals never include +C.
Example:
โซ_1^3 2x dx = [x^2]_1^3 = 3^2 - 1^2 = 9 - 1 = 8
The definite integral $\int_a^b f(x) \, dx$ equals the signed area between the curve and the $x$-axis:
Example: $\int_0^{2\pi} \sin x \, dx = [-\cos x]_0^{2\pi} = (-\cos 2\pi) - (-\cos 0) = (-1) - (-1) = 0$.
The positive area from $0$ to $\pi$ exactly cancels the negative area from $\pi$ to $2\pi$. If you want the total physical area (always positive), compute $\int_0^{\pi} \sin x \, dx + |\int_{\pi}^{2\pi} \sin x \, dx| = 2 + 2 = 4$.
Show y = sin(x) from 0 to 2ฯ. Shade the region from 0 to ฯ above the x-axis in blue (positive). Shade the region from ฯ to 2ฯ below the x-axis in red (negative). Mark that the two regions have equal area but opposite sign, giving total integral = 0. Add labels showing the positive lobe (+2) and negative lobe (-2).
Evaluate $\int_0^2 (e^x + x) \, dx$.
The numerical value.
โซ(e^x + x) dx = e^x + x^2/2
[e^x + x^2/2]_0^2
= (e^2 + 2^2/2) - (e^0 + 0)
= (e^2 + 2) - (1 + 0)
= e^2 + 1
โ 7.389 + 1 = 8.389
$e^2 + 1 \approx 8.39$.
Evaluate $\int_1^e \frac{1}{x} \, dx$.
Answer:
$[\ln x]_1^e = \ln e - \ln 1 = 1 - 0 = 1$.
$\int_a^b f(x) \, dx = F(b) - F(a)$
Definite integrals give numbers
$\int_a^b = -\int_b^a$
$\int_a^b + \int_b^c = \int_a^c$
The distance is $\int_0^3 2t \, dt = [t^2]_0^3 = 9 - 0 = 9$ metres, which is more than 6 metres. Since the car is speeding up ($v = 2t$ increases with time), it travels faster in the later part of the interval. The average speed is greater than the initial speed (0 m/s) but the simple multiplication $2 \times 3 = 6$ uses the speed at $t = 1$ (2 m/s), which underestimates the total distance.
5 random questions from a replayable lesson bank โ feedback shown immediately
Short Answer
8. Evaluate $\int_0^2 (3x^2 + 4x - 1) \, dx$. Show all working. 3 MARKS
9. Evaluate $\int_1^3 (e^x + \frac{1}{x}) \, dx$. Give your answer in exact form. 3 MARKS
10. A car's velocity is given by $v(t) = 12 - 2t$ m/s for $0 \leq t \leq 8$. Find the total distance travelled in the first 8 seconds. Explain why you cannot simply evaluate $\int_0^8 (12 - 2t) \, dx$ to find distance, and describe what the definite integral $\int_0^8 (12 - 2t) \, dt$ actually represents in this context. 3 MARKS
1. $[\frac{x^3}{3}]_0^3 = 9 - 0 = 9$
2. $[e^x]_1^2 = e^2 - e$
3. $[\ln x]_1^4 = \ln 4 - \ln 1 = \ln 4 = 2\ln 2$
4. $[x^3 + x^2 + x]_0^1 = 1 + 1 + 1 = 3$
1. The positive area from $0$ to $\pi$ equals the negative area from $\pi$ to $2\pi$, so they cancel.
2. $s = \int_0^5 6t \, dt = [3t^2]_0^5 = 75$ metres.
3. $[\frac{x^4}{4}]_{-1}^1 = \frac{1}{4} - \frac{1}{4} = 0$. $x^3$ is an odd function, so symmetric areas cancel.
Q8 (3 marks): $F(x) = x^3 + 2x^2 - x$ [1]. $[F(x)]_0^2 = (8 + 8 - 2) - (0)$ [1] $= 14$ [1].
Q9 (3 marks): $F(x) = e^x + \ln x$ [1]. $[F(x)]_1^3 = (e^3 + \ln 3) - (e^1 + \ln 1)$ [1] $= e^3 - e + \ln 3$ [1].
Q10 (3 marks): $v(t) = 12 - 2t = 0$ when $t = 6$ [0.5]. For $0 \leq t \leq 6$: $v \geq 0$; for $6 \leq t \leq 8$: $v \leq 0$ [0.25]. Distance = $\int_0^6 (12-2t) \, dt + |\int_6^8 (12-2t) \, dt|$ [0.5] $= [12t - t^2]_0^6 + |[12t - t^2]_6^8|$ [0.25] $= (72 - 36) + |(96 - 64) - (72 - 36)|$ [0.25] $= 36 + |32 - 36|$ [0.25] $= 36 + 4 = 40$ metres [0.5]. The definite integral $\int_0^8 (12-2t) \, dt = 36 - 4 = 32$ represents the displacement (net change in position), not distance [0.25]. After $t = 6$, the car moves backward, so the displacement counts this negatively while distance counts it positively [0.25].
Climb platforms using definite integrals, properties, and area interpretation. Pool: lesson 4.
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