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Pure · Integration

Chapter 1 · 4

The idea

Areas under parametric curves

Finding the area under a curve given parametrically as ∫ y (dx/dt) dt with the limits in t — and combining it with tangents/normals to find regions bounded by the curve, a line and an axis.

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Pure · Integration

Areas under parametric curves

Finding the area under a curve given parametrically as ∫ y (dx/dt) dt with the limits in t — and combining it with tangents/normals to find regions bounded by the curve, a line and an axis.

Why it works

Substituting t into the area integral

The area under a curve is ∫y dx\displaystyle\int y\,dx. When the curve is given parametrically — x=f(t)x = f(t), y=g(t)y = g(t) — both yy and dxdx need to be put in terms of tt before you can integrate:

∫y dx=∫t1t2y dxdt dt\int y\,dx = \int_{t_1}^{t_2} y\,\frac{dx}{dt}\,dt123452468xy

A first computation

Take the curve x=t2, y=4tx = t^2,\ y = 4t (for t≥0t \ge 0). To find the area between it and the xx-axis from x=0x = 0 to x=4x = 4, note these are t=0t = 0 and t=2t = 2. With dxdt=2t\dfrac{dx}{dt} = 2t,

∫y dxdt dt=∫024t⋅2t dt=∫028t2 dt=[83t3]02=643.\int y\,\frac{dx}{dt}\,dt = \int_0^2 4t\cdot 2t\,dt = \int_0^2 8t^2\,dt = \left[\tfrac83 t^3\right]_0^2 = \frac{64}{3}.

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