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MathematicsCalculus

Parametric integration

Parametric integration finds the area under a curve whose x and y are both given in terms of a parameter, usually t. The rule is area = ∫ y (dx/dt) dt: multiply y by the derivative of x with respect to t, then integrate with respect to t, using limits that are values of t, not x. It lets you find areas without ever writing the curve as y in terms of x.

The parametric integration formula

For a curve with x = f(t) and y = g(t), the area between the curve and the x-axis, from x = a to x = b, is:

\int_a^b y\ dx = \int_{t_1}^{t_2} y\frac{dx}{dt}\ dtwhere t₁ and t₂ are the values of t when x = a and x = b

It comes from the chain rule, or more simply from swapping dx for its value in terms of dt. Since dx/dt is the rate at which x changes as t changes, dx = (dx/dt) dt. Put that into ∫ y dx and every part is now in terms of t.

How to integrate parametric equations

  1. Differentiate x with respect to t to find dx/dt.
  2. Change the limits. Solve x = a and x = b for t. These are your new limits.
  3. Write y × dx/dt in terms of t, and simplify.
  4. Integrate with respect to t and substitute the t limits.
  5. Check the sign. An area should be positive. If x decreases as t increases, the answer comes out negative: take its size, or swap the limits.

Parametric integration examples

Example 1: a simple area, checked two ways

A curve has x = t² and y = 2t, for t ≥ 0. Find the area under it between x = 1 and x = 9.

dx/dt = 2t. When x = 1, t = 1; when x = 9, t = 3.

\int_1^3 2t \times 2t\ dt = \int_1^3 4t^2\ dt = \left[\frac{4t^3}{3}\right]_1^3 = \frac{108}{3} - \frac{4}{3} = \frac{104}{3}

The area is 104/3, about 34.7. Check: eliminating t gives y = 2√x, and the integral of 2√x from 1 to 9 is (4/3)(27 − 1) = 104/3. The same answer, as it should be.

Example 2: the area of an ellipse

An ellipse has parametric equations x = 3 cos t, y = 2 sin t. Find the area of the quarter in the first quadrant, then the whole ellipse.

The quarter runs from x = 0 to x = 3. dx/dt = −3 sin t. When x = 0, t = π/2; when x = 3, t = 0.

\int_{\pi/2}^{0} 2\sin t \times (-3\sin t)\ dt = \int_0^{\pi/2} 6\sin^2 t\ dt

Swapping the limits flipped the sign and made it positive. Now use the double angle formula, sin²t = (1 − cos 2t)/2:

\int_0^{\pi/2} 3(1 - \cos 2t)\ dt = \left[3t - \frac{3}{2}\sin 2t\right]_0^{\pi/2} = \frac{3\pi}{2}

The quarter has area 3π/2, so the whole ellipse has area 4 × 3π/2 = 6π. That matches the general formula for an ellipse, πab, with a = 3 and b = 2.

y = 2\sqrt{1-\frac{{ x}^{2}}{9}}
-3.5-3-2.5-2-1.5-1-0.50.511.522.533.50.511.52t = π/2: (0, 2)t = 0: (3, 0)
Drag to pan, scroll to zoom. x from -3.5 to 3.5, y from -0.4 to 2.4.

Here x is across and y is up: the curve is the top half of the ellipse, written as y in terms of x. The area from x = 0 to x = 3, under the right-hand part of the curve, is the quarter worked out above. As t goes from 0 to π/2, the point moves from (3, 0) to (0, 2), right to left, which is why x decreases and the limits had to be swapped.

Example 3: integration by parts in t

A curve has x = eᵗ and y = t. Find the area under it from x = 1 to x = e².

dx/dt = eᵗ. When x = 1, t = 0; when x = e², t = 2. So the area is the integral of teᵗ from 0 to 2, which needs integration by parts:

\int_0^2 te^t\ dt = \left[te^t - e^t\right]_0^2 = (2e^2 - e^2) - (0 - 1) = e^2 + 1

The area is e² + 1, about 8.39. The curve is y = ln x in disguise, and integrating ln x from 1 to e² gives the same answer.

This practice from the Calculus course evaluates definite integrals, the last step of every parametric area question. Evaluate the integral, then press Mark to check it in the course.

Evaluate a definite integral, giving a numeric or fraction answer.

Press Mark to have your answer marked in the Calculus course.

When the curve crosses the axis

As with ordinary integration, an area below the x-axis comes out negative. If a parametric curve dips below the axis, find each part separately: work out the t values where y = 0, integrate over each section, and add the sizes of the areas. The Integral Calculator is a quick way to check each part numerically.

Common mistakes

  • Using x values as the limits. After the substitution, everything is in t, so the limits must be t values.
  • Forgetting dx/dt. The integral is ∫ y (dx/dt) dt, not ∫ y dt.
  • Using dy/dt by mistake. It is the derivative of x that goes in, because the area is ∫ y dx.
  • Ignoring a negative answer. If x decreases as t increases, the integral is negative. The area is its size.
  • Not simplifying before integrating. Products like sin²t need an identity first.

Is parametric integration in the formula booklet?

No. Neither AQA's nor Edexcel's A-level Maths formulae booklet, checked in November 2026, gives the rule ∫ y (dx/dt) dt, so you need to know it. Both booklets do give integration by parts, which parametric questions often need, and Edexcel's lists the integrals of several trig functions. The double angle formulae you need for sin² and cos² are not in either booklet. Our guide to double angle formulae covers them.

Parametric integration questions with answers

1. A curve has x = t + 1 and y = t². Find the area under it between t = 0 and t = 2.
dx/dt = 1, so the area is ∫ t² dt from 0 to 2 = [t³/3] = 8/3
2. A curve has x = 2t and y = 1/t, for t > 0. Find the area under it from x = 2 to x = 6.
dx/dt = 2. When x = 2, t = 1; when x = 6, t = 3.
∫ (1/t) × 2 dt from 1 to 3 = [2 ln t] = 2 ln 3, about 2.20
3. A circle has x = cos t, y = sin t. Use parametric integration to find the area of the quarter in the first quadrant.
dx/dt = −sin t. From x = 0 (t = π/2) to x = 1 (t = 0): ∫ sin t (−sin t) dt from π/2 to 0 = ∫ sin²t dt from 0 to π/2.
= [t/2 − sin 2t/4] from 0 to π/2 = π/4
4. A curve has x = t² and y = t³, for t ≥ 0. Find the area under it from x = 0 to x = 4.
dx/dt = 2t. When x = 0, t = 0; when x = 4, t = 2.
∫ t³ × 2t dt = ∫ 2t⁴ dt from 0 to 2 = [2t⁵/5] = 64/5

Practise integration in a short timed paper

A-level Maths: Calculus gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across differentiation and integration, including parametric equations. In the US, the same practice opens as AP Calculus AB.

Paper resets in

Common questions

What is the formula for parametric integration?
Area = ∫ y (dx/dt) dt, with the limits written as values of t.
Why do you multiply by dx/dt?
Because the area is ∫ y dx, and dx = (dx/dt) dt. Multiplying by dx/dt turns the integral into one with respect to t.
Do you need to change the limits?
Yes. The original limits are x values. Find the t values that give those x values, and use them.
Why is my parametric area negative?
Probably because x decreases as t increases between your limits. Take the size of the answer, or swap the limits.
Can you just convert to Cartesian form instead?
Sometimes, but eliminating t can give an awkward equation, and many parametric curves have no simple y = f(x) form. The parametric method always works.

Revision cards

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