# Parametric Integrals

# Parametric Integrals

**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.

**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

A Level**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]**

x=3t+4

\dfrac{dx}{dt}=3

y=t^{2}

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

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

A Level

**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**.

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}

A Level## 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}

**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}

**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}

## Related Topics

#### Parametric Equations

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