Parametric Integrals

When dealing with parametric equations, integrals become more complicated. We cannot just do $\int y \, dx$ when we don’t have $y$ written in terms of $x$ . Instead, we must use the chain rule to get an integral in terms of the parameter. Then, if it is a definite integral, we must convert the limits to fit the new integration.

Make sure you are happy with the following topics before continuing.

Parametric Equations

Product

A Level Maths Predicted Papers 2025

117

From: £19.99

The MME A level maths predicted papers are an excellent way to practise, using authentic exam style questions that are unique to our papers. Our examiners have studied A level maths past papers to develop predicted A level maths exam questions in an authentic exam format. The profit from every pack is reinvested into making free content on MME, which benefits millions of learners across the country.

View Product

Product

A-A* A Level Maths Practice Papers

5

£19.99

The MME A-A* A level maths practice papers are excellent for those top achieving students to practise for their exams, using authentic exam style questions that are unique to our practice papers. Our content experts have studied A level maths past papers and specifications to develop specific A-A* A level maths exam questions in an authentic exam style. The profit from every pack is reinvested into making free content on MME, which benefits millions of learners across the country.

View Product

The Chain Rule

Recall: The Chain Rule.

$\dfrac{dy}{dx}=\dfrac{dy}{dz}\dfrac{dz}{dx}$

If we have parametric equations and $y$ isn’t written in terms of $x$ , but instead it is written in terms of $t$ say, then we can use the chain rule to show that $dx=\dfrac{dx}{dt} \, dt$ for a parameter $t$ , and since we have $x$ in terms of $t$ we can get $\dfrac{dx}{dt}$ in terms of $t$ , and we already have $y$ in terms of $t$ , so our integral can be written as:

${\LARGE \int} y \, dx={\LARGE \int} y \, \dfrac{dx}{dt} \, dt$

Limit Conversion

If we have a definite integral $\int^{b}_{a}y \, dx$ , then we cannot just take our limits $a$ and $b$ and put them on our new integral in terms of $t$ , because they are limits with respect to $x$ .

Instead, we need to convert them.

This means that the lower limit on the integral in terms of $t$ is the $t$ value that gives $x=a$ , and the upper limit on the integral in terms of $t$ is the $t$ value that gives $x=b$ .

With these converted limits we can find the value of the definite integral.

Example 1: Using the Chain Rule

A parametric equation is $x=3t+4$ and $y=t^{2}$ . Find $\int y \, dx$ in terms of $t$ .

[2 marks]

\int y \, dx={\LARGE \int} y\dfrac{dx}{dt} \, dt

x=3t+4

\dfrac{dx}{dt}=3

y=t^{2}

\int y \, dx=\int 3t^{2} \, dt

\int y \, dx=t^{3}+c

Example 2: Definite Integrals

A parametric curve is defined by $y=t^{3}+3t$ , $x=t^{2}+4t+4$ , for $t>-3$ . Find $\int^{4}_{0}y \, dx$ .

[3 marks]

First convert limits.

x=t^{2}+4t+4

First limit: $x=0$

$t^{2}+4t+4=0$

$(t+2)^{2}=0$

$t=-2$

Second limit: $x=4$

$t^{2}+4t+4=4$

$t^{2}+4t=0$

$t(t+4)=0$

$t=0$ or $t=-4$

$t=-4$ not in range

t=0

$\int^{4}_{0} y \, dx={\LARGE\int}^{0}_{-2} \, y\dfrac{dx}{dt} \, dt$

$x=t^{2}+4t+4$

$\dfrac{dx}{dt}=2t+4$

$y=t^{3}+3t$

$\begin{aligned}\int^{4}_{0} y \, dx&=\int^{0}_{-2}(t^{3}+3t)(2t+4) \, dt\\[1.2em]&=\int^{0}_{-2}\left( 2t^{4}+4t^{3}+6t^{2}+12t\right) dt\\[1.2em]&=\left[\dfrac{2}{5}t^{5}+t^{4}+2t^{3}+6t^{2}\right]^{0}_{-2}\\[1.2em]&=\left( \dfrac{2}{5}\times0^{5}\right) +0^{4}+\left( 2\times0^{3}\right) +\left( 6\times0^{2}\right)-\left( \dfrac{2}{5}\times(-2)^{5}\right) -(-2)^{4} - 2(- 2)^{3} - 6( -2)^{2} \\[1.2em]&=\left( \dfrac{2}{5}\times32\right) -16+\left( 2\times8\right) -\left( 6\times4\right) \\[1.2em]&=\dfrac{64}{5}-16+16-24\\[1.2em]&=-\dfrac{56}{5}\end{aligned}$

Parametric Integrals Example Questions

Question 1: A curve has parametric equation $x=t^{2}+2$ , $y=t^{3}+4t^{2}+4t+3$ . Show that $\int y \, dx=\int\left( 2t^{4}+8t^{3}+8t^{2}+6t\right) dt$

[2 marks]

$\int y \, dx={\LARGE \int} y\dfrac{dx}{dt}dt$

$x=t^{2}+2$

$\dfrac{dx}{dt}=2t$

$y=t^{3}+4t^{2}+4t+3$

$\begin{aligned}\int y \, dx&=\int\left( 2t(t^{3}+4t^{2}+4t+3)\right) \, dt \\[1.2em]&=\int\left( 2t^{4}+8t^{3}+8t^{2}+6t\right) \, dt \end{aligned}$

Save your answers with

Gold Standard Education

Sign Up Now

Question 2: Find $\int y \, dx$ in terms of $t$ , where $y=t^{-\frac{1}{2}}$ and $x=4t^{\frac{5}{9}}$

[2 marks]

$\int y \, dx={\LARGE \int} y\dfrac{dx}{dt}dt$

$x=4t^{\frac{5}{9}}$

$\dfrac{dx}{dt}=\dfrac{20}{9}t^{-\frac{4}{9}}$

$y=t^{-\frac{1}{2}}$

$\begin{aligned}\int ydx&=\int \left( t^{-\frac{1}{2}}\times\dfrac{20}{9}t^{-\frac{4}{9}}\right) dt\\[1.2em]&=\int \dfrac{20}{9}t^{-\frac{17}{18}}dt\\[1.2em]&=\left( \dfrac{20}{9}\div\dfrac{1}{18}\right) t^{\frac{1}{18}}+c\\[1.2em]&=40t^{\frac{1}{18}}+c\end{aligned}$

Save your answers with

Gold Standard Education

Sign Up Now

Question 3: A parametric curve is defined by $x=2t+4$ , $y=t+6$ . Find $\int^{6}_{0}y \, dx$

[3 marks]

First find the new limits.

Upper limit $x=6$

$2t+4=6$

$2t=2$

$t=1$

Lower limit $x=0$

$2t+4=0$

$2t=-4$

$t=-2$

Thus:

$\int^{6}_{0}y \, dx={\LARGE \int}^{1}_{-2}y\dfrac{dx}{dt} \, dt$

$x=2t+4$

$\dfrac{dx}{dt}=2$

$y=t+6$

$\begin{aligned}\int^{6}_{0}y \, dx&=\int^{1}_{-2}2(t+6) \, dt\\[1.2em]&=\int^{1}_{-2}\left( 2t+12\right) \, dt\\[1.2em]&=[t^{2}+12t]^{1}_{-2}\\[1.2em]&=1^{2}+\left( 12\times1\right) -(-2)^{2}-\left( 12\times(-2)\right) \\[1.2em]&=1+12-4+24\\[1.2em]&=33\end{aligned}$

Save your answers with

Gold Standard Education

Sign Up Now