Integration Practice Questions

Q: How do I integrate \dfrac{1}{x^{2}} or a square root?

Rewrite them as powers first. 1x2=x−2\dfrac{1}{x^{2}}=x^{-2}x21=x−2 and x=x1/2\sqrt{x}=x^{1/2}x=x1/2, then apply ∫axn dx=an+1xn+1+c\int ax^{n}\,dx=\dfrac{a}{n+1}x^{n+1}+c∫axndx=n+1axn+1+c as usual. The only value that fails is n=−1n=-1n=−1, which is beyond this chapter.

Q: What is the difference between the area formula and the volume formula?

The area under a curve is ∫aby dx\int_{a}^{b} y\,dx∫abydx, while the volume when that region is rotated about the xxx-axis is π∫aby2 dx\pi\int_{a}^{b} y^{2}\,dxπ∫aby2dx. The volume squares yyy and multiplies by π\piπ; mixing the two up is a common slip.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from a plain indefinite integral to the volume of a solid of revolution. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, the rule you are using, a clear substitution of the powers or limits, then the final answer.

Never forget the arbitrary constant $c$ on an indefinite integral, and resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, a correctly quoted rule still earns credit even when the arithmetic slips, so always write the integration rule before you substitute, and check an indefinite integral by differentiating it back.

Six practice questions

Q1[3 marks]

Find $\displaystyle\int (6x^{2}-4x+5)\,dx$ .

Show worked solution

Integrate term by term with the power rule $\int ax^{n}\,dx=\dfrac{a}{n+1}x^{n+1}+c$ , and add a single arbitrary constant $c$ at the end:

\int (6x^{2}-4x+5)\,dx=\frac{6}{3}x^{3}-\frac{4}{2}x^{2}+5x+c

=2x^{3}-2x^{2}+5x+c

Answer

The integral is $2x^{3}-2x^{2}+5x+c$ . Check by differentiating: $\dfrac{d}{dx}\left(2x^{3}-2x^{2}+5x+c\right)=6x^{2}-4x+5$ , which is the original integrand.

Q2[3 marks]

Find $\displaystyle\int \left(4x^{3}-\frac{1}{x^{2}}\right)dx$ .

Show worked solution

Rewrite $\dfrac{1}{x^{2}}$ as $x^{-2}$ so the power rule applies, then integrate each term. Note that $n+1=-1$ for the second term:

\int \left(4x^{3}-x^{-2}\right)dx=\frac{4}{4}x^{4}-\frac{x^{-1}}{-1}+c

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

Answer

The integral is $x^{4}+\dfrac{1}{x}+c$ . Check: $\dfrac{d}{dx}\left(x^{4}+x^{-1}\right)=4x^{3}-x^{-2}=4x^{3}-\dfrac{1}{x^{2}}$ .

Q3[3 marks]

Find $\displaystyle\int (2x+1)^{3}\,dx$ .

Show worked solution

For a linear bracket raised to a power use $\int (ax+b)^{n}\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ . Here $a=2$ and $n=3$ , so $a(n+1)=2\times 4=8$ :

\int (2x+1)^{3}\,dx=\frac{(2x+1)^{4}}{2(4)}+c=\frac{(2x+1)^{4}}{8}+c

Answer

The integral is $\dfrac{(2x+1)^{4}}{8}+c$ . Check: $\dfrac{d}{dx}\dfrac{(2x+1)^{4}}{8}=\dfrac{4(2x+1)^{3}\times 2}{8}=(2x+1)^{3}$ , using the chain rule.

Q4[4 marks]

The gradient function of a curve is $\dfrac{dy}{dx}=3x^{2}-4x$ , and the curve passes through the point $(2,5)$ . Find the equation of the curve.

Show worked solution

The curve is the integral of its gradient function, so integrate to recover $y$ , keeping the constant $c$ :

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

Use the point $(2,5)$ to fix $c$ : substitute $x=2$ and $y=5$ :

5=(2)^{3}-2(2)^{2}+c=8-8+c \;\Rightarrow\; c=5

y=x^{3}-2x^{2}+5

Answer

The equation is $y=x^{3}-2x^{2}+5$ . Check: at $x=2$ , $y=8-8+5=5$ , and differentiating gives $\dfrac{dy}{dx}=3x^{2}-4x$ , both as required.

Q5[4 marks]

Find the area of the region bounded by the curve $y=x^{2}+1$ , the $x$ -axis, and the lines $x=0$ and $x=2$ .

Show worked solution

The area between a curve that lies above the $x$ -axis and the $x$ -axis is $\displaystyle\int_{a}^{b} y\,dx$ . Here the region runs from $x=0$ to $x=2$ :

\text{Area}=\int_{0}^{2}(x^{2}+1)\,dx=\left[\frac{x^{3}}{3}+x\right]_{0}^{2}

Substitute the upper limit, then subtract the value at the lower limit:

=\left(\frac{8}{3}+2\right)-(0)=\frac{8}{3}+\frac{6}{3}=\frac{14}{3}

Answer

The area is $\dfrac{14}{3}$ (about $4.67$ ) square units. Because $y=x^{2}+1>0$ throughout $0\le x\le 2$ , the whole region lies above the axis and the integral is positive.

Q6[5 marks]

The region bounded by the line $y=2x$ , the $x$ -axis and the line $x=3$ is rotated $360^{\circ}$ about the $x$ -axis. Find the volume of the solid generated, in terms of $\pi$ .

Show worked solution

For a region rotated $360^{\circ}$ about the $x$ -axis the volume is $V=\pi\displaystyle\int_{a}^{b} y^{2}\,dx$ . With $y=2x$ we have $y^{2}=4x^{2}$ , and the region runs from $x=0$ to $x=3$ :

V=\pi\int_{0}^{3}(2x)^{2}\,dx=\pi\int_{0}^{3}4x^{2}\,dx

=\pi\left[\frac{4x^{3}}{3}\right]_{0}^{3}=\pi\left(\frac{4(27)}{3}-0\right)=36\pi

Answer

The volume is $36\pi$ cubic units. Check with the cone formula: the solid is a cone of radius $r=2(3)=6$ and height $h=3$ , so $V=\tfrac{1}{3}\pi r^{2}h=\tfrac{1}{3}\pi(36)(3)=36\pi$ , which agrees.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct rule, the right substitution of powers or limits, and a clean final statement.

Method mark: did you quote the correct rule, $\int ax^{n}\,dx=\dfrac{a}{n+1}x^{n+1}+c$ , the $(ax+b)^{n}$ rule, the area formula $\int y\,dx$ , or the volume formula $\pi\int y^{2}\,dx$ ?
Constant mark: on an indefinite integral, did you write $+c$ ? Leaving it out costs a mark even when everything else is correct.
Limits mark: on a definite integral, did you evaluate the antiderivative at the upper limit and subtract the value at the lower limit, in that order?
Answer mark: is the final value or expression stated clearly, and does an indefinite integral survive a check by differentiating it back to the integrand?
For an area or volume, confirm the region lies above the axis (or split it where it crosses) so the integral is not accidentally negative.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a missing $+c$ , a power rule applied to $(2x+1)^{3}$ without dividing by the extra factor of $2$ , or limits substituted in the wrong order, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each rule. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

Do I always need to write

+c

Yes, on every indefinite integral. Integration reverses differentiation, and any constant differentiates to zero, so the constant $c$ must be written to describe the whole family of answers.

On a definite integral the constant cancels when you subtract, so you do not need it there.

How do I integrate

\dfrac{1}{x^{2}}

or a square root?

Rewrite them as powers first. $\dfrac{1}{x^{2}}=x^{-2}$ and $\sqrt{x}=x^{1/2}$ , then apply $\int ax^{n}\,dx=\dfrac{a}{n+1}x^{n+1}+c$ as usual.

The only value that fails is $n=-1$ , which is beyond this chapter.

What is the difference between the area formula and the volume formula?

The area under a curve is $\int_{a}^{b} y\,dx$ , while the volume when that region is rotated about the $x$ -axis is $\pi\int_{a}^{b} y^{2}\,dx$ . The volume squares $y$ and multiplies by $\pi$ ; mixing the two up is a common slip.

How can I check an integration answer quickly?

Differentiate it. If your integral is correct, differentiating it returns the original integrand.

This single check catches most sign and coefficient errors before you lose the mark.

Source:SRC-DSKP-EN

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

Integration, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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