# 6.4 Solving Problems by Indefinite Integration

F.5 Mathematics Module 1 · Chapter 6 Indefinite Integration · Lesson 2

From a slope or a rate of change back to the curve or the quantity: integrate, then use the given point, slope or starting value to fix the constant.

## Key points

- **Geometrical meaning**: indefinite integration is the reverse process of differentiation, so it finds the equation of a curve when the **slope of the tangent ／ 切線的斜率** at any point $(x, y)$ is given.
- **A family of curves**: $\frac{dy}{dx} = 2x$ gives $y = \int 2x\,dx = x^2 + C$, one curve for each value of the **arbitrary constant ／ 任意常數** $C$, each a translation of $y = x^2$ upwards or downwards. Extra information fixes $C$: through $(1, -3)$ means $-3 = 1^2 + C$, so $y = x^2 - 4$.
- **Second derivative given**: integrate twice, and each integration brings its own constant, $C_1$ then $C_2$. Find $C_1$ from the slope condition before integrating again (Example 2).
- **A tangent line** $y = mx + k$ at $P$ gives the slope $m$: solve $\frac{dy}{dx} = m$ to locate $P$, then use $P$ on the curve to find $C$.
- **Rate of change ／ 變化率**: if $R = \frac{dQ}{dt}$, then $Q = \int R\,dt$. $R$ is positive when $Q$ is increasing and negative when $Q$ is decreasing.
- **Starting values hidden in words**: "total … generated since the launch" → $Q(0) = 0$; "released from rest" → $v = 0$ when $t = 0$; "fully charged" → $E(0) = 100$ (%); "from a height of 210 m" → $s = 210$ when $t = 0$.
- **Motion**: integrate the acceleration for $v$, then $v$ for the **displacement ／ 位移** $s$ (Example 5: $a = -10$, $v = -10t$, $s = 210 - 5t^2$, ground when $s = 0$ at $t = 6.48$ s; the negative root is rejected).
- **Claims**: work out the value, compare it with the claim, then write "∴ The claim is agreed." or "disagreed." "After a very long time" means $t \to \infty$: use $e^{-kt} \to 0$ for $k > 0$.

## Must know

| The question gives … | It means … | Use it to … |
|---|---|---|
| $\dfrac{dy}{dx} = f(x)$ | the slope of the tangent at any point $(x, y)$ | integrate: $y = \int f(x)\,dx = F(x) + C$, a family of curves |
| the curve passes through $(a, b)$ | $y = b$ when $x = a$ | find $C$ from $y = F(x) + C$ |
| $\dfrac{d^2y}{dx^2} = g(x)$ | integrate twice; each integration brings its own constant | $\dfrac{dy}{dx} = \int g(x)\,dx + C_1$, then integrate again with $+\,C_2$ |
| the slope of the tangent at $x = a$ is $m$ | $\dfrac{dy}{dx} = m$ when $x = a$ | find $C_1$ |
| the line $y = mx + k$ touches the curve at $P$ | slope at $P$ is $m$; $P$ lies on the line and on the curve | solve $f(x) = m$ for $P$ first, then find $C$ |
| $(a, b)$ is a maximum / minimum point | $\dfrac{dy}{dx} = 0$ and $y = b$ when $x = a$ | two conditions for two constants |
| a rate of change $\dfrac{dQ}{dt} = R(t)$ | $Q$ is found by integrating the rate | $Q = \int R(t)\,dt$; a known value such as $Q(0)$ gives $C$ |
| "after a very long time" | let $t \to \infty$ | use $e^{-kt} \to 0$ for $k > 0$ |

Every integration needs its $+\,C$; every constant needs one condition from the question.

## Common mistakes

- ✗ $y = x^3 + Cx + C$ after integrating $\frac{d^2y}{dx^2} = 6x$ twice. ✓ Two integrations, two different constants: $\frac{dy}{dx} = 3x^2 + C_1$, $y = x^3 + C_1x + C_2$.
- ✗ Curve through $(1, 2)$, slope 4 there: $2 = 3(1)^2 + C_1$, so $C_1 = -1$. ✓ The point belongs to $y$, the slope to $\frac{dy}{dx}$: $4 = 3(1)^2 + C_1$ gives $C_1 = 1$, then $2 = 1 + 1 + C_2$ gives $C_2 = 0$ and $y = x^3 + x$.
- ✗ Tangent $y = -2x + 5$ at $P$: reading $P$ off the line, or using $k = 5$ as $C$. ✓ The line gives the slope $m = -2$: solve $\frac{dy}{dx} = -2$ for the $x$ of $P$, put it into the line for $y$, then find $C$ from $P$.
- ✗ Distance travelled in the first 6 s $= s(6) - s(0)$. ✓ That is the displacement. When $v$ changes sign, split the time where $v = 0$ and add the distances of each part.
- ✗ Total revenue in the first year: $R'(1)$, or $R'(1) \times 1$ (rate × time). ✓ The rate is not constant: $R(t) = \int R'(t)\,dt$ with $R(0) = 0$, and the first year's total is $R(1) - R(0)$.

## Quick check

1. The slope of the tangent to a curve at any point $(x, y)$ is $\frac{dy}{dx} = 4x - 3$ and the curve passes through $(2, 5)$. Find the equation of the curve.
2. At any point $(x, y)$ of a curve, $\frac{d^2y}{dx^2} = 12x$. The curve passes through $(1, 5)$ and the slope of the tangent there is 8. Find the equation of the curve.
3. Water flows into a tank at a rate of $\frac{dV}{dt} = 6e^{0.5t}$ litres per minute, where $V$ litres is the volume after $t$ minutes and the tank is empty at $t = 0$. Express $V$ in terms of $t$ and find $V$ when $t = 2$ (correct to the nearest litre).
4. A ball is thrown vertically upwards from the ground with velocity 20 m/s and its acceleration is always $-10$ m/s². Find $v$ in terms of $t$ and the greatest height reached.
5. An object moves with $v(t) = 2t - 4$ m/s for $0 \le t \le 4$, and $s(0) = 3$. Find $s(t)$ and the total distance travelled in the first 4 s.

## Answers

1. $y = 2x^2 - 3x + C$; $5 = 8 - 6 + C$, $C = 3$: $y = 2x^2 - 3x + 3$.
2. $\frac{dy}{dx} = 6x^2 + C_1$, $8 = 6 + C_1$, $C_1 = 2$; $y = 2x^3 + 2x + C_2$, $5 = 2 + 2 + C_2$, $C_2 = 1$: $y = 2x^3 + 2x + 1$.
3. $V = 12e^{0.5t} + C$, $0 = 12 + C$: $V = 12e^{0.5t} - 12$; $V(2) = 12e - 12 \approx 21$ litres.
4. $v = -10t + 20$; $v = 0$ at $t = 2$; $s = 20t - 5t^2$ (as $s = 0$ at $t = 0$), so the greatest height is $s(2) = 20$ m.
5. $s = t^2 - 4t + 3$; $v = 0$ at $t = 2$: $s(0) = 3$, $s(2) = -1$, $s(4) = 3$; distance $= 4 + 4 = 8$ m (the displacement is 0).

## Did you know?

Integrating a constant acceleration twice gives a fall distance of $5t^2$, growing with the square of the time (height $s = 210 - 5t^2$ in Example 5): the law Galileo found in the early 1600s by rolling balls down ramps, decades before calculus existed.