Integration Worked Examples (Medium)

Q: How do I find the equation of a curve from its gradient function?

Integrate the gradient function to get yyy with a constant +c+c+c, then substitute the given point to solve for ccc. Without the point you can only find the family of curves, not the single one asked for.

What these examples cover

These medium Integration examples move from mechanics to problems that ask you to combine two steps. You will recover the equation of a curve from its gradient function and a point, evaluate a definite integral that first needs the integrand rewritten as separate powers, and find the area of a region bounded by a curve and the x-axis.

Each still uses small, clean numbers, but now the marks reward reading the question carefully and setting the work out in order. Cover the solution, attempt each fully on paper, then check line by line.

When your answer differs, trace back to the exact step where the two solutions diverge, that is where the understanding is.

Worked examples

Work through all three. Attempt each fully before you read the matching solution, and notice how each one starts by turning the question into a clean integral before any arithmetic begins.

Q1[4 marks]

The gradient function of a curve is $\dfrac{dy}{dx}=3x^{2}-6x+2$ , and the curve passes through the point $(1,4)$ . Find the equation of the curve.

Show worked solution

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

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

The value of $c$ is fixed by the point the curve passes through. Substitute $(x,y)=(1,4)$ :

4=(1)^{3}-3(1)^{2}+2(1)+c=1-3+2+c=c

So $c=4$ , and the equation of the curve is:

y=x^{3}-3x^{2}+2x+4

Answer

The curve is $y=x^{3}-3x^{2}+2x+4$ . Check the point: at $x=1$ , $y=1-3+2+4=4$ , which matches $(1,4)$ ; and differentiating returns $3x^{2}-6x+2$ , the given gradient function.

Q2[4 marks]

Evaluate $\int_{1}^{4}\dfrac{x^{2}-2}{\sqrt{x}}\,dx$ .

Show worked solution

A quotient like this cannot be integrated as it stands. Split it into separate terms over $\sqrt{x}=x^{1/2}$ , then write each as a single power using $\dfrac{x^{a}}{x^{b}}=x^{a-b}$ :

\frac{x^{2}-2}{\sqrt{x}}=\frac{x^{2}}{x^{1/2}}-\frac{2}{x^{1/2}}=x^{3/2}-2x^{-1/2}

Integrate each power, raising the index by one:

\int (x^{3/2}-2x^{-1/2})\,dx=\frac{x^{5/2}}{5/2}-\frac{2x^{1/2}}{1/2}=\frac{2}{5}x^{5/2}-4x^{1/2}

Now apply the limits $x=4$ then $x=1$ . Note $4^{5/2}=(\sqrt{4})^{5}=2^{5}=32$ and $4^{1/2}=2$ :

\left[\frac{2}{5}x^{5/2}-4x^{1/2}\right]_{1}^{4}=\left(\frac{2}{5}(32)-4(2)\right)-\left(\frac{2}{5}(1)-4(1)\right)

=\left(\frac{64}{5}-8\right)-\left(\frac{2}{5}-4\right)=\frac{24}{5}-\left(-\frac{18}{5}\right)=\frac{42}{5}

Answer

$\int_{1}^{4}\dfrac{x^{2}-2}{\sqrt{x}}\,dx=\dfrac{42}{5}$ (that is $8.4$ ). The move that unlocks the whole question is splitting the fraction into separate powers before integrating, the integral of a quotient is not the quotient of the integrals.

Q3[5 marks]

The curve $y=4x-x^{2}$ meets the x-axis at two points. Find the area of the region enclosed between the curve and the x-axis.

Show worked solution

First find where the curve meets the x-axis by setting $y=0$ and factorising:

4x-x^{2}=0 \;\Rightarrow\; x(4-x)=0 \;\Rightarrow\; x=0 \text{ or } x=4

These are the limits of integration. Between $x=0$ and $x=4$ the curve lies above the x-axis, so the area is the definite integral of $y$ :

\text{Area}=\int_{0}^{4}(4x-x^{2})\,dx=\left[2x^{2}-\frac{x^{3}}{3}\right]_{0}^{4}

Substitute the limits; the lower limit $x=0$ contributes nothing:

=\left(2(4)^{2}-\frac{(4)^{3}}{3}\right)-0=32-\frac{64}{3}=\frac{96-64}{3}=\frac{32}{3}

Answer

The area is $\dfrac{32}{3}$ square units (about $10.67$ ). Finding the intercepts first is essential, they give the correct limits, and skipping that step is the usual reason an area answer comes out wrong.

Q4[3 marks]

Evaluate $\int_{0}^{1}(2x+1)^{3}\,dx$ .

Show worked solution

This is a composite power of a linear expression, so use the rule $\displaystyle\int (ax+b)^{n}\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ rather than expanding the bracket term by term. Here $a=2$ and $n=3$ :

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

Apply the limits $x=1$ then $x=0$ :

\left[\frac{(2x+1)^{4}}{8}\right]_{0}^{1}=\frac{(3)^{4}}{8}-\frac{(1)^{4}}{8}=\frac{81}{8}-\frac{1}{8}=\frac{80}{8}=10

Answer

$\int_{0}^{1}(2x+1)^{3}\,dx=10$ . Using the linear rule avoids expanding $(2x+1)^{3}$ into four separate terms, always check that the coefficient $a$ sits in the denominator alongside $n+1$ .

Q5[4 marks]

The region bounded by the curve $y=x^{2}$ , the x-axis, and the line $x=2$ is rotated $360°$ about the x-axis. Find the volume of the solid formed.

Show worked solution

When a region under a curve is rotated fully about the x-axis, the volume generated is $\displaystyle V=\pi\int_{a}^{b}y^{2}\,dx$ . Here $y^{2}=(x^{2})^{2}=x^{4}$ , and the region runs from $x=0$ to $x=2$ :

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

Integrate, then substitute the limits:

V=\pi\left[\frac{x^{5}}{5}\right]_{0}^{2}=\pi\left(\frac{32}{5}-0\right)=\frac{32\pi}{5}

Answer

The volume is $\dfrac{32\pi}{5}$ cubic units (about $20.1$ ). Remember to square $y$ before integrating, using $y$ instead of $y^{2}$ is the most common slip in a volume-of-revolution question.

Q6[4 marks]

The curve $y=x^{2}$ lies in the first quadrant. Find the area of the region bounded by the curve, the y-axis, and the line $y=4$ .

Show worked solution

This region is bounded by the y-axis, so integrate with respect to $y$ using $\displaystyle\text{Area}=\int x\,dy$ . Rewrite $y=x^{2}$ with $x$ as the subject, taking the positive root since the curve lies in the first quadrant:

y=x^{2} \;\Rightarrow\; x=\sqrt{y}=y^{1/2}

The region runs from $y=0$ to $y=4$ . Integrate $x$ with respect to $y$ over these limits:

\text{Area}=\int_{0}^{4}y^{1/2}\,dy=\left[\frac{2}{3}y^{3/2}\right]_{0}^{4}

Substitute the limits, using $4^{3/2}=(\sqrt{4})^{3}=2^{3}=8$ :

=\frac{2}{3}(8)-0=\frac{16}{3}

Answer

The area is $\dfrac{16}{3}$ square units (about $5.33$ ). Whenever a region is bounded by the y-axis rather than the x-axis, rewrite the curve with $x$ as the subject and integrate with respect to $y$ .

Q7[4 marks]

Find the area of the region enclosed between the line $y=x+2$ and the curve $y=x^{2}$ .

Show worked solution

First find where the line and curve intersect, by setting the two expressions for $y$ equal:

x^{2}=x+2 \;\Rightarrow\; x^{2}-x-2=0 \;\Rightarrow\; (x-2)(x+1)=0 \;\Rightarrow\; x=-1 \text{ or } x=2

These are the limits of integration. Between $x=-1$ and $x=2$ the line lies above the curve, so the area is the integral of (line $-$ curve):

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

Substitute the limits:

=\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{9}{2}

Answer

The enclosed area is $\dfrac{9}{2}$ square units ( $4.5$ ). For two curves, always integrate (upper $-$ lower), checking which graph is on top at a point between the intersections (e.g. $x=0$ ) avoids a sign error.

Q8[4 marks]

A particle moves along a straight line and passes through a fixed point $O$ when $t=0$ . Its velocity, $v \text{ m s}^{-1}$ , at time $t$ seconds is given by $v=6t-3t^{2}$ .

Find the displacement of the particle from $O$ when $t=2$ .

Show worked solution

Displacement is the integral of velocity, so integrate $v$ with respect to $t$ , keeping the constant of integration $c$ :

s=\int (6t-3t^{2})\,dt=3t^{2}-t^{3}+c

The particle passes through $O$ when $t=0$ , which means $s=0$ at $t=0$ . Substitute to find $c$ :

0=3(0)^{2}-(0)^{3}+c \;\Rightarrow\; c=0

So $s=3t^{2}-t^{3}$ . Substitute $t=2$ to find the displacement:

s=3(2)^{2}-(2)^{3}=12-8=4

Answer

The displacement from $O$ when $t=2$ is $4$ m. The phrase 'passes through $O$ when $t=0$ ' is what fixes $c=0$ , without it, the displacement formula would only be known up to a constant.

Notice the shared shape of all three: read the question, turn it into a definite or indefinite integral with the right limits or constant, integrate cleanly, and only then substitute. When you separate the setting-up from the arithmetic, medium Integration questions become steady, predictable marks.

Key method points

These three examples rehearse the reasoning that medium Integration questions reward. Keep the following points in mind as you practise more.

To recover a curve from $\dfrac{dy}{dx}$ , integrate, then use a given point to find the constant $c$ .
Split a single fraction into separate power terms before integrating: $\dfrac{x^{2}-2}{\sqrt{x}}=x^{3/2}-2x^{-1/2}$ .
Evaluate powers of the limits carefully: $4^{5/2}=(\sqrt{4})^{5}=2^{5}=32$ .
For an area bounded by a curve and the x-axis, find the x-intercepts first, they are your limits.
An area is a positive quantity; if a definite integral comes out negative, the region lies below the axis.
Set the working out line by line so method marks are secure under analytic marking.

How a teacher helps

On medium questions the marks usually turn on setting up rather than on the integration itself, forgetting to use the point to pin down $c$ , trying to integrate a fraction without splitting it, or reaching for an area without first finding the limits. In a one-to-one lesson our teacher slows down exactly at the setting-up stage, so the integral you write is the right one before any arithmetic starts.

Because our teachers are experienced, you learn a repeatable method, not a one-off trick. Lessons are taught in English, while SPM papers are set in both Malay and English.

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Frequently asked questions

How do I find the equation of a curve from its gradient function?

Integrate the gradient function to get $y$ with a constant $+c$ , then substitute the given point to solve for $c$ . Without the point you can only find the family of curves, not the single one asked for.

Why can't I integrate

\dfrac{x^{2}-2}{\sqrt{x}}

directly?

There is no rule for the integral of a quotient. Split it into separate terms first, $x^{3/2}-2x^{-1/2}$ , so the power rule applies to each term on its own.

Why do I need the x-intercepts before finding an area?

They are the limits of the region. The area between a curve and the x-axis runs from one intercept to the next, so integrating between the wrong values gives the wrong area.

What if my area comes out negative?

A negative value means the region is below the x-axis over those limits. Area is positive, so take the modulus (magnitude) of the integral for that part.

Source:SRC-DSKP-EN

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

Integration, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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