A water tank is being filled at a rate that changes with time: $\frac{dV}{dt} = 3t^2$ litres per minute. How much water enters the tank in the first 4 minutes? To answer this, we need to integrate power functions fluently — handling positive powers, negative powers, and fractional powers with confidence. This lesson sharpens the power rule into a reliable tool for any situation.
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What is $\int \frac{1}{\sqrt{x}} \, dx$? Predict before calculating — think about rewriting $\frac{1}{\sqrt{x}}$ as a power of $x$.
Core Content
The power rule works for any power except $n = -1$:
∫x^n dx = x^(n+1)/(n+1) + C, n ≠ -1
Rewriting before integrating:
Common pattern: When integrating $x^{-n}$ where $n > 1$, the answer is a negative power (or fraction) of $x$.
When a function is not written as a sum of powers, expand it first:
Example: ∫(x + 1)^2 dx = ∫(x^2 + 2x + 1) dx = x^3/3 + x^2 + x + C
Example: ∫(x^2 + 1)/x dx = ∫(x + 1/x) dx = x^2/2 + ln|x| + C
Example: ∫x(x + 3) dx = ∫(x^2 + 3x) dx = x^3/3 + 3x^2/2 + C
Rule: Divide by $x$ before integrating if the numerator has higher degree. Expand brackets if present. Simplify fractions.
Find $\int \frac{x^3 + 2x}{x} \, dx$.
The antiderivative.
Divide each term by x:
(x^3 + 2x)/x = x^2 + 2
∫(x^2 + 2) dx = x^3/3 + 2x + C
$\frac{x^3}{3} + 2x + C$.
Find $\int \frac{x^2 - 3x + 2}{x} \, dx$.
Answer:
$\frac{x^2}{2} - 3x + 2\ln|x| + C$.
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$
$\sqrt{x} = x^{1/2}$
$\frac{1}{x^n} = x^{-n}$
Expand, divide, rewrite — then integrate
Differentiate your answer
$\int \frac{1}{\sqrt{x}} \, dx = \int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} + C = 2\sqrt{x} + C$. The key step is rewriting $\frac{1}{\sqrt{x}}$ as $x^{-1/2}$. Many students forget the negative sign in the exponent or make an arithmetic error with $\frac{1}{1/2} = 2$.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find $\int \frac{x^3 + 4x^2 - x}{x^2} \, dx$. Show all working. 3 MARKS
9. Find $\int (\sqrt{x} + \frac{1}{\sqrt{x}})^2 \, dx$. Expand first, then integrate term by term. 3 MARKS
10. A water tank fills at rate $\frac{dV}{dt} = 6\sqrt{t}$ litres per minute. Find the total volume of water that enters the tank in the first 9 minutes. Then find how long it takes for the total volume to reach 144 litres. Explain why the time to reach 144 L is more than twice the time to reach 72 L. 3 MARKS
1. $\frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$
2. $\int x^{2/3} \, dx = \frac{x^{5/3}}{5/3} + C = \frac{3}{5}x^{5/3} + C$
3. $\int (x^2 - 2x^{-1}) \, dx = \frac{x^3}{3} - 2\ln|x| + C$
4. $\frac{x^{5/2}}{5/2} + C = \frac{2}{5}x^{5/2} + C$
1. Power rule gives $\frac{x^0}{0}$ which is undefined. Correct: $\int \frac{1}{x} \, dx = \ln|x| + C$.
2. $\int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} + C = 2x^{1/2} + C = 2\sqrt{x} + C$.
3. $V = \int_0^4 4t^{-1/2} \, dt = 8t^{1/2}\big|_0^4 = 8(2) = 16$ litres.
Q8 (3 marks): $\frac{x^3 + 4x^2 - x}{x^2} = x + 4 - \frac{1}{x}$ [1]. $\int (x + 4 - \frac{1}{x}) \, dx = \frac{x^2}{2} + 4x - \ln|x| + C$ [2].
Q9 (3 marks): $(\sqrt{x} + \frac{1}{\sqrt{x}})^2 = x + 2 + \frac{1}{x}$ [1]. $\int (x + 2 + \frac{1}{x}) \, dx = \frac{x^2}{2} + 2x + \ln|x| + C$ [2].
Q10 (3 marks): $V = \int 6\sqrt{t} \, dt = 6 \cdot \frac{t^{3/2}}{3/2} + C = 4t^{3/2} + C$ [0.5]. At $t = 0$, $V = 0$, so $C = 0$ [0.25]. In 9 minutes: $V = 4(9)^{3/2} = 4(27) = 108$ litres [0.5]. Set $4t^{3/2} = 144$ [0.25]: $t^{3/2} = 36$, so $t = 36^{2/3} = (36^{1/3})^2 \approx 3.30^2 \approx 10.9$ minutes [0.5]. Time to 72 L: $4t^{3/2} = 72$, $t^{3/2} = 18$, $t = 18^{2/3} \approx 6.87$ minutes [0.25]. Time to 144 L (10.9 min) is more than twice time to 72 L (6.87 min) because the fill rate decreases as $\sqrt{t}$ grows more slowly than $t$ — the tank fills at a decelerating rate [0.75].
Climb platforms using fractional powers, negative powers, and expanding before integrating. Pool: lesson 2.
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