Integration Worked Examples (Easy)

Q: Why do I divide by a for \int (ax+b)^{n}\,dx?

Because differentiating (ax+b)n+1(ax+b)^{n+1}(ax+b)n+1 brings out an extra factor of aaa by the chain rule. Dividing by aaa as well as by n+1n+1n+1 cancels it, so that differentiating your answer returns the original bracket.

Q: Do I have to rewrite fractions before integrating?

Yes, it is the safest route. Writing 2x2\dfrac{2}{x^{2}}x22 as 2x−22x^{-2}2x−2 lets the power rule apply directly, so you avoid the common slip of trying to integrate a fraction as it stands.

What these examples cover

These easy Integration examples build the everyday moves the whole chapter rests on: integrating a polynomial term by term and remembering the constant of integration, rewriting a reciprocal or negative power before integrating, using the shortcut for a linear bracket raised to a power $(ax+b)^{n}$ , and evaluating a definite integral between two limits. Each uses small, clean numbers so you can follow every line without a calculator getting in the way.

Cover the solution, attempt the question in full on paper, and only then check line by line against our working. Where your answer differs, find the exact step where the two solutions part company, that single line is usually where the real learning is.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, raise the index by one, divide by the new index, and keep track of the constant, runs through every one.

Q1[3 marks]

Find $\int (6x^{2}-4x+3)\,dx$ .

Show worked solution

Integrate term by term. For each power term use $\int ax^{n}\,dx=\dfrac{a}{n+1}x^{n+1}+c$ , raising the index by one and dividing by the new index; the constant $3$ integrates to $3x$ .

\int (6x^{2}-4x+3)\,dx=\frac{6x^{3}}{3}-\frac{4x^{2}}{2}+3x+c

Simplify each coefficient:

=2x^{3}-2x^{2}+3x+c

Answer

$\int (6x^{2}-4x+3)\,dx=2x^{3}-2x^{2}+3x+c$ . Never drop the $+c$ on an indefinite integral.

Check by differentiating: $\frac{d}{dx}(2x^{3}-2x^{2}+3x+c)=6x^{2}-4x+3$ , the original integrand.

Q2[3 marks]

Find $\int \left(x^{3}-\dfrac{2}{x^{2}}\right)dx$ .

Show worked solution

The power rule for integration needs every term written as a power of $x$ . Rewrite the reciprocal: $\dfrac{2}{x^{2}}=2x^{-2}$ .

\int \left(x^{3}-\frac{2}{x^{2}}\right)dx=\int (x^{3}-2x^{-2})\,dx

Integrate each term, raising each index by one. For $-2x^{-2}$ the new index is $-2+1=-1$ :

=\frac{x^{4}}{4}-\frac{2x^{-1}}{-1}+c=\frac{x^{4}}{4}+2x^{-1}+c

Write the negative power back as a fraction so the answer matches the form of the question:

=\frac{x^{4}}{4}+\frac{2}{x}+c

Answer

$\int \left(x^{3}-\dfrac{2}{x^{2}}\right)dx=\dfrac{x^{4}}{4}+\dfrac{2}{x}+c$ . Rewriting $\dfrac{2}{x^{2}}$ as $2x^{-2}$ first is what lets the power rule apply; dividing by $n+1=-1$ flips the sign, which is why the term becomes $+\dfrac{2}{x}$ .

Q3[3 marks]

Find $\int (2x+1)^{4}\,dx$ .

Show worked solution

For a linear bracket raised to a power, use $\int (ax+b)^{n}\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ : raise the power by one, then divide by the new power and by $a$ , the coefficient of $x$ . Here $a=2$ and $n=4$ .

\int (2x+1)^{4}\,dx=\frac{(2x+1)^{5}}{2\times 5}+c=\frac{(2x+1)^{5}}{10}+c

Answer

$\int (2x+1)^{4}\,dx=\dfrac{(2x+1)^{5}}{10}+c$ . Check by differentiating: $\frac{d}{dx}\dfrac{(2x+1)^{5}}{10}=\dfrac{5(2x+1)^{4}\times 2}{10}=(2x+1)^{4}$ .

The extra division by $a=2$ is the step students most often miss.

Q4[4 marks]

Evaluate $\int_{1}^{2}(3x^{2}+2)\,dx$ .

Show worked solution

First integrate to get the antiderivative, no $+c$ is needed for a definite integral, because it cancels when you subtract. Then apply the limits with $\int_{a}^{b}f(x)\,dx=\big[F(x)\big]_{a}^{b}=F(b)-F(a)$ .

\int_{1}^{2}(3x^{2}+2)\,dx=\big[x^{3}+2x\big]_{1}^{2}

Substitute the upper limit $x=2$ , then the lower limit $x=1$ , and subtract:

=\big(2^{3}+2(2)\big)-\big(1^{3}+2(1)\big)=(8+4)-(1+2)=12-3=9

Answer

$\int_{1}^{2}(3x^{2}+2)\,dx=9$ . Always substitute the top limit first and subtract the bottom; reversing the order flips the sign of your answer.

Q5[3 marks]

The gradient function of a curve is $\dfrac{dy}{dx}=4x-3$ , and the curve passes through the point $(1,5)$ . Find the equation of the curve.

Show worked solution

Integrate the gradient function to get $y$ in terms of $x$ , with a constant of integration $c$ still to be found.

y=\int (4x-3)\,dx=2x^{2}-3x+c

Substitute the given point $x=1,\,y=5$ to find $c$ .

5=2(1)^{2}-3(1)+c=-1+c\ \Rightarrow\ c=6

Answer

The equation of the curve is $y=2x^{2}-3x+6$ . A gradient function alone gives a whole family of curves; the given point picks out this one member by fixing $c$ .

Q6[2 marks]

Find $\int (6\sqrt{x}+1)\,dx$ .

Show worked solution

Write $\sqrt{x}$ as a fractional power, $x^{1/2}$ , so the power rule applies directly.

\int (6\sqrt{x}+1)\,dx=\int (6x^{1/2}+1)\,dx

Integrate each term, raising the index of $x^{1/2}$ by one to $\tfrac{3}{2}$ .

=\frac{6x^{3/2}}{3/2}+x+c=4x^{3/2}+x+c

Answer

$\int (6\sqrt{x}+1)\,dx=4x^{3/2}+x+c$ . Check by differentiating: $\frac{d}{dx}(4x^{3/2}+x+c)=4\times\frac{3}{2}x^{1/2}+1=6\sqrt{x}+1$ .

Q7[3 marks]

Find $\int \dfrac{x^{3}+4x}{x}\,dx$ .

Show worked solution

Before integrating, divide each term in the numerator by $x$ so the fraction becomes a sum of powers of $x$ .

\frac{x^{3}+4x}{x}=\frac{x^{3}}{x}+\frac{4x}{x}=x^{2}+4

Now integrate the simplified expression term by term.

\int (x^{2}+4)\,dx=\frac{x^{3}}{3}+4x+c

Answer

$\int \dfrac{x^{3}+4x}{x}\,dx=\dfrac{x^{3}}{3}+4x+c$ . Simplifying the fraction first avoids trying to integrate a quotient directly, which the power rule cannot handle.

Q8[2 marks]

Given that $\int_{2}^{5} f(x)\,dx=6$ and $\int_{5}^{8} f(x)\,dx=3$ , find the value of $\int_{2}^{8} f(x)\,dx$ .

Show worked solution

Use the property that two integrals over adjoining intervals add to give the integral over the combined interval.

\int_{2}^{8} f(x)\,dx=\int_{2}^{5} f(x)\,dx+\int_{5}^{8} f(x)\,dx

=6+3=9

Answer

$\int_{2}^{8} f(x)\,dx=9$ . No actual antiderivative is needed here, the shared limit $x=5$ lets the two given integrals be added directly.

Different as these four look, the routine underneath is the same: get every term into power form, integrate carefully one term at a time, and keep the constant or the limits in clear view. That steadiness turns Integration into a dependable source of marks in both papers.

Key method points

These four examples rehearse the skills that open almost every Integration question in Add Math. Keep the following points in mind as you practise more.

Integrate a power term with $\int ax^{n}\,dx=\dfrac{a}{n+1}x^{n+1}+c$ : raise the index by one, divide by the new index.
Every indefinite integral needs a constant of integration $+c$ ; a definite integral does not.
Rewrite reciprocals and roots as powers, $\dfrac{2}{x^{2}}=2x^{-2}$ , $\sqrt{x}=x^{1/2}$ , before integrating.
For a linear bracket, $\int (ax+b)^{n}\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ ; do not forget to divide by $a$ .
For a definite integral, evaluate $F(b)-F(a)$ , substituting the upper limit first.
Because marking is analytic, a correct integration line can still earn method marks even if the final arithmetic slips.

How a teacher helps

When a student drops a mark on questions like these, it is usually a small, fixable habit, a dropped $+c$ , a reciprocal left un-rewritten, or forgetting to divide by $a$ in the linear-bracket rule. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains the why behind each step, not just the what. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

When do I need the

+c

On every indefinite integral, one with no limits. It stands for the family of curves that all share the same gradient function.

A definite integral, with a top and bottom limit, does not need it because the constant cancels when you subtract.

Why do I divide by

a

for

\int (ax+b)^{n}\,dx

Because differentiating $(ax+b)^{n+1}$ brings out an extra factor of $a$ by the chain rule. Dividing by $a$ as well as by $n+1$ cancels it, so that differentiating your answer returns the original bracket.

Do I have to rewrite fractions before integrating?

Yes, it is the safest route. Writing $\dfrac{2}{x^{2}}$ as $2x^{-2}$ lets the power rule apply directly, so you avoid the common slip of trying to integrate a fraction as it stands.

Which limit do I substitute first in a definite integral?

The upper (top) limit, then subtract the value from the lower (bottom) limit: $F(b)-F(a)$ . Doing it the other way round flips the sign and loses the accuracy mark.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Integration, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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