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.

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Parametric Equations

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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]

$int y , dx={LARGE int} ydfrac{dx}{dt} , dt$

$x=3t+4$

$dfrac{dx}{dt}=3$

$y=t^{2}$

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

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

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

$t=0$

$int^{4}_{0} y , dx={LARGEint}^{0}_{-2} , ydfrac{dx}{dt} , dt$

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

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

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

$begin{aligned}int^{4}_{0} y , dx&=int^{0}_{-2}(t^{3}+3t)(2t+4) , dt[1.2em]&=int^{0}_{-2}left( 2t^{4}+4t^{3}+6t^{2}+12tright) dt[1.2em]&=left[dfrac{2}{5}t^{5}+t^{4}+2t^{3}+6t^{2}right]^{0}_{-2}[1.2em]&=left( dfrac{2}{5}times0^{5}right) +0^{4}+left( 2times0^{3}right) +left( 6times0^{2}right)-left( dfrac{2}{5}times(-2)^{5}right) -(-2)^{4} – 2(- 2)^{3} – 6( -2)^{2} [1.2em]&=left( dfrac{2}{5}times32right) -16+left( 2times8right) -left( 6times4right) [1.2em]&=dfrac{64}{5}-16+16-24[1.2em]&=-dfrac{56}{5}end{aligned}$

Parametric Integrals Example Questions

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

[2 marks]

$int y , dx={LARGE int} ydfrac{dx}{dt}dt$

$x=t^{2}+2$

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

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

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

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

[2 marks]

$int y , dx={LARGE int} ydfrac{dx}{dt}dt$

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

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

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

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

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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:

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

$x=2t+4$

$dfrac{dx}{dt}=2$

$y=t+6$

$begin{aligned}int^{6}_{0}y , dx&=int^{1}_{-2}2(t+6) , dt[1.2em]&=int^{1}_{-2}left( 2t+12right) , dt[1.2em]&=[t^{2}+12t]^{1}_{-2}[1.2em]&=1^{2}+left( 12times1right) -(-2)^{2}-left( 12times(-2)right) [1.2em]&=1+12-4+24[1.2em]&=33end{aligned}$