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Module 6 · L2 of 15 ~35 min ⚡ +95 XP available

Integrating Power Functions

A water tank fills at a rate that changes with time: $\frac{dV}{dt} = 3t^2$ litres per minute. How much water enters 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.

Today's hook — What is $\int \frac{1}{\sqrt{x}} \, dx$? Predict before calculating — think about rewriting $\frac{1}{\sqrt{x}}$ as a power of $x$. The answer will surprise you if you try it without rewriting first.

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

+5 XP warm-up

What is $\int \dfrac{1}{\sqrt{x}} \, dx$? Predict before calculating — think about rewriting $\frac{1}{\sqrt{x}}$ as a power of $x$. What exponent does $\frac{1}{\sqrt{x}}$ have?

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The two moves that unlock any power function

+5 XP to read

Before integrating, always rewrite the function as a sum of powers. Two moves cover everything:

Move 1 — Rewrite roots as fractional powers:

$\sqrt{x} = x^{1/2}$ $\sqrt[3]{x} = x^{1/3}$ $\frac{1}{x^n} = x^{-n}$ $\frac{1}{\sqrt{x}} = x^{-1/2}$

Move 2 — Expand or divide first:

$(x+1)^2 = x^2 + 2x + 1$ before integrating. $\frac{x^3 + 2x}{x} = x^2 + 2$ before integrating.

$\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + C,\quad n \neq -1$

$\int \sqrt{x} \, dx$

$= \int x^{1/2} \, dx = \dfrac{x^{3/2}}{3/2} + C = \dfrac{2}{3}x^{3/2} + C$

$\int \frac{1}{\sqrt{x}} \, dx$

$= \int x^{-1/2} \, dx = \dfrac{x^{1/2}}{1/2} + C = 2\sqrt{x} + C$

$\int \frac{1}{x^2} \, dx$

$= \int x^{-2} \, dx = \dfrac{x^{-1}}{-1} + C = -\dfrac{1}{x} + C$

Water flow and tank filling. A tank fills at rate $\frac{dV}{dt} = \frac{5}{\sqrt{t}}$ litres per minute. How much water enters in the first 9 minutes? $V = \int_0^9 5t^{-1/2} \, dt = 5 \cdot 2t^{1/2}\big|_0^9 = 10(\sqrt{9} - 0) = 30$ litres. The decreasing rate (water pressure drops as tank fills) means the flow slows over time, but integration captures the total precisely.

What you'll master

Know

Key facts

Power rule for all $n \neq -1$
How to rewrite roots and reciprocals as powers
Expanding brackets before integrating

Understand

Concepts

Why $n = -1$ is a special case
How algebraic manipulation prepares functions for integration
The connection between power rules for differentiation and integration

Can do

Skills

Integrate functions involving roots and reciprocals
Expand brackets and integrate term by term
Simplify fractions before integrating

Key terms

Fractional power$x^{p/q}$ where $p$ and $q$ are integers; represents roots, e.g. $x^{1/2} = \sqrt{x}$.

Negative power$x^{-n} = \frac{1}{x^n}$; represents reciprocals of powers.

Expand then integrateExpand brackets or simplify fractions before applying the power rule to each term.

Divide by $x$If a fraction has $x$ in the denominator, divide each numerator term before integrating.

$n = -1$ exception$\int x^{-1} \, dx = \ln|x| + C$ — the power rule breaks down at $n = -1$.

Check by differentiatingDifferentiate your answer — if you recover the original integrand, the answer is correct.

Fractional and negative powers

core concept

The power rule works for any power except $n = -1$. The key step is always rewriting before applying the rule.

$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$

Full worked pattern — $\int \frac{1}{\sqrt{x}} \, dx$:

Rewrite: $\frac{1}{\sqrt{x}} = x^{-1/2}$
Apply power rule: $\int x^{-1/2} \, dx = \frac{x^{-1/2 + 1}}{-1/2 + 1} = \frac{x^{1/2}}{1/2}$
Simplify: $\frac{x^{1/2}}{1/2} = 2x^{1/2} = 2\sqrt{x}$
Write final answer: $2\sqrt{x} + C$

Common pattern: When integrating $x^{-n}$ where $n > 1$, the answer is a negative power (or fraction) of $x$. When integrating $x^{-1/2}$, $\frac{1}{-1/2 + 1} = \frac{1}{1/2} = 2$ — a common arithmetic slip to watch for.

$\sqrt{x} = x^{1/2}$ $\frac{1}{x^n} = x^{-n}$ $\sqrt[n]{x^m} = x^{m/n}$; $\int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} + C$

Pause — copy the rewriting rules $\sqrt{x} = x^{1/2}$, $\frac{1}{x^n} = x^{-n}$, $\sqrt[n]{x^m} = x^{m/n}$, and the result $\int\sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C$ into your book.

Did you get this? True or false: to find $\int \frac{1}{\sqrt{x}} \, dx$, the first step is to rewrite $\frac{1}{\sqrt{x}}$ as $x^{-1/2}$ before applying the power rule.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · NEGATIVE AND FRACTIONAL POWERS

Find $\int x^{-3} \, dx$ and $\int x^{3/2} \, dx$.

$\int x^{-3} \, dx = \dfrac{x^{-3+1}}{-3+1} = \dfrac{x^{-2}}{-2} = -\dfrac{1}{2x^2} + C$

Add 1 to power ($-3 \to -2$), divide by new power ($-2$). Rewrite with positive denominator.

PROBLEM 2 · EXPANDING BEFORE INTEGRATING

Find $\int (x + 1)^2 \, dx$.

$(x + 1)^2 = x^2 + 2x + 1$

Expand the bracket — we cannot integrate $(x+1)^2$ directly using the power rule (that would need the chain rule for differentiation in reverse).

PROBLEM 3 · DIVIDE BY x FIRST

Find $\int \dfrac{x^3 + 2x}{x} \, dx$.

$\dfrac{x^3 + 2x}{x} = x^2 + 2$

Divide each term in the numerator by $x$. Now both terms are plain powers.

Quick check: What is the first step to find $\int \dfrac{x^4 - 2x}{x^2} \, dx$?

Common errors · the 3 traps that cost marks

Trap 01

Forgetting to rewrite before integrating

Writing $\int \sqrt{x} \, dx$ and trying to apply the power rule to "$\sqrt{x}$" directly won't work. You must rewrite as $x^{1/2}$ first. Every root and reciprocal must become a power before the rule applies.

Trap 02

Dividing by a fraction error

$\frac{x^{3/2}}{3/2}$ does NOT equal $\frac{x^{3/2}}{1.5}$ left unsimplified. Multiply by the reciprocal: $\frac{1}{3/2} = \frac{2}{3}$. Always simplify: $\frac{2}{3}x^{3/2} + C$.

Trap 03

Integrating $(x+1)^2$ as $\frac{(x+1)^3}{3}$

$\int (x+1)^2 \, dx \neq \frac{(x+1)^3}{3} + C$. That would only be correct if the chain rule applied perfectly (it doesn't for non-linear brackets). Always expand first.

Copy into your book. Write out the rewriting rules ($\sqrt{x}$, $\frac{1}{x^n}$, $\sqrt[n]{x^m}$) and the two strategy steps (expand brackets, divide by $x$) with one example of each.

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Quick-fire practice · 4 antiderivatives

Find $\int x^{-3} \, dx$

Find $\int \sqrt[3]{x^2} \, dx$ (rewrite as $x^{2/3}$)

Find $\int \dfrac{x^4 - 2x}{x^2} \, dx$ (divide first)

Find $\int x^{3/2} \, dx$

Fill in the blank: $\int \dfrac{1}{\sqrt{x}} \, dx = \int x^{-1/2} \, dx = \dfrac{x^{1/2}}{1/2} + C = \boxed{?}\sqrt{x} + C$. What is the missing coefficient?

Real world: tank filling at a variable rate

A tank fills at $\frac{dV}{dt} = \frac{4}{\sqrt{t}}$ L/min. Find the total water in the first 4 minutes.

$V = \int_0^4 4t^{-1/2} \, dt$

$= 4 \cdot \frac{t^{1/2}}{1/2}\bigg|_0^4 = 8t^{1/2}\bigg|_0^4 = 8(\sqrt{4} - \sqrt{0}) = 8(2) = 16$ litres.

Now try this: show that $\int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} + C$ by rewriting and applying the power rule step by step.

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Match the function to its antiderivative. Which is the correct antiderivative of $\sqrt[3]{x^2} = x^{2/3}$?

Revisit your thinking

Earlier you were asked: what is $\int \frac{1}{\sqrt{x}} \, dx$? The answer is $2\sqrt{x} + C$. 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 $\frac{1}{1/2} = 2$. Were you close?

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. Find $\int \dfrac{x^3 + 4x^2 - x}{x^2} \, dx$. Show all working. (3 marks)

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ApplyBand 43 marks

Q2. Find $\int \left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)^2 \, dx$. Expand first, then integrate term by term. (3 marks)

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AnalyseBand 53 marks

Q3. A water tank fills at rate $\dfrac{dV}{dt} = 6\sqrt{t}$ 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)

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Comprehensive answers (click to reveal)

Drill 1: $-\frac{1}{2x^2} + C$ 2: $\frac{3}{5}x^{5/3} + C$ 3: $\frac{x^3}{3} - 2\ln|x| + C$ 4: $\frac{2}{5}x^{5/2} + C$

Q1 (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].

Q2 (3 marks): $\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = x + 2 + \frac{1}{x}$ [1]. $\int (x + 2 + \frac{1}{x}) \, dx = \frac{x^2}{2} + 2x + \ln|x| + C$ [2].

Q3 (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 min: $V = 4(9)^{3/2} = 4(27) = 108$ litres [0.5]. Set $4t^{3/2} = 144$: $t^{3/2} = 36$, so $t = 36^{2/3} \approx 10.9$ min [0.5]. Time to 72 L: $4t^{3/2} = 72$, $t^{3/2} = 18$, $t \approx 6.87$ min [0.25]. Time to 144 L (10.9 min) is more than twice time to 72 L (6.87 min) because the fill rate $6\sqrt{t}$ grows more slowly than $t$ — the tank fills at a decelerating rate [0.75].

Boss battle · The Power Master

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms using fractional powers, negative powers, and expanding before integrating. Pool: lesson 2.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module overview · Maths Advanced · Checkpoint 1