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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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Orient to power functions

Work through the visible teaching and complete each embedded check.

Worksheets

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

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Recall, your gut answer first

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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

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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.

Convert roots and reciprocals to powers, then apply the power rule.

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

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

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

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

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

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

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

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

Build the rewriting language

Work through the visible teaching and complete each embedded check.

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.

Rewrite before integrating

Work through the visible teaching and complete each embedded check.

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.

Work through power integrals

Work through the visible teaching and complete each embedded check.

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

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

Expanding is one safe route and here it is the quickest. It is NOT the only one: $\dfrac{(x+1)^3}{3} + C$ is also correct, because differentiating it gives $(x+1)^2$ — the inner derivative is 1, so no extra factor appears. The factor only bites when the inner coefficient is not 1, as in $\int (2x+1)^2\,dx = \dfrac{(2x+1)^3}{6} + C$, where dividing by that 2 is what the reverse chain rule is for.

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 $\int \dfrac{x^4 - 2x}{x^2} \, dx$?

Avoid rewriting traps

Work through the visible teaching and complete each embedded check.

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

Carrying $\frac{(x+1)^3}{3}$ over to $(2x+1)^2$

$\int (x+1)^2 \, dx = \frac{(x+1)^3}{3} + C$ IS correct — differentiate it and the inner derivative is 1, so no extra factor appears (Problem 2). The trap is reusing that shortcut when the inner derivative is not 1: $\int (2x+1)^2 \, dx = \frac{(2x+1)^3}{6} + C$, not $\frac{(2x+1)^3}{3} + C$, because you must divide by the inner derivative 2. Unsure? Expand first, or check by differentiating your answer.

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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Apply and reflect

Work through the visible teaching and complete each embedded check.

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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Complete the mixed practice

Complete the question-bank round and short-answer practice, then compare your responses with the comprehensive answers.

Multiple choice

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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 less than twice the time to 72 L (6.87 min), since $2 \times 6.87 = 13.74 > 10.9$: doubling the volume multiplies the time by $2^{2/3} \approx 1.59$. The fill rate $6\sqrt{t}$ increases with $t$, so the tank fills at an accelerating rate [0.75].

Review and complete

Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.

Retrieve the lesson before you leave

Rewrite a root and a reciprocal using indices.
State why the power rule excludes an exponent of negative one.
Explain why expanding or dividing first can make an integral solvable.

Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.

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