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Areas, parametric curves and the limit of a sum questions
Three upgrades to area-finding: the gap between two curves surrenders to a single integral of top minus bottom, curves given parametrically integrate without ever leaving the parameter, and underneath it all sits the definition, an integral as the limit of a sum of ever-thinner strips.
7 original questions · 32 marks · the areas, parametric curves and the limit of a sum notes · Integration
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Find the area of the region bounded by the curve y = 4 − x2, the x-axis and the y-axis, in the first quadrant.
Worked answer
Setting y = 0 gives 4 − x2 = 0, so the right-hand limit is x = 2, and the y-axis supplies x = 0. Then ∫02 (4 − x2) dx = [4x − x3/3] from 0 to 2 = 8 − 8/3 = 16/3. B1 for the upper limit, M1 for the integration, A1 for 16/3. Reading the upper limit off a rough sketch rather than solving y = 0 is the usual route to x = 4 and a wrong answer.Find the area enclosed between the line y = x and the curve y = x2.
Worked answer
Solve x = x2 for the limits, giving x = 0 and x = 1. Between them the line rides above the parabola, so the area is ∫01 (x − x2) dx = 1/2 − 1/3 = 1/6. M1 for equating to find the limits, A1 for x = 0 and x = 1, M1 for integrating the difference, A1 for 1/6. Top minus bottom, integrated across the enclosed stretch. Subtract the other way round and you get −1/6; a negative area earns no accuracy mark, however tidy the integration.Find the area enclosed between the curves y = 9 − x2 and y = x2 + 1.
Worked answer
Equating gives 9 − x2 = x2 + 1, so x2 = 4 and the curves cross at x = ±2. The vertical gap is (9 − x2) − (x2 + 1) = 8 − 2x2, and ∫−22 (8 − 2x2) dx = [8x − 2x3/3] from −2 to 2 = 32/3 − (−32/3) = 64/3. M1 A1 for the limits x = ±2, M1 for the difference of the curves, M1 for the integration, A1 for 64/3. One integral of the height difference handles both boundaries at once. Integrating each curve separately and subtracting works too, but it doubles the chances of a slip at the negative limit.The curve C has parametric equations x = t3, y = 3t2, t ≥ 0. Find the exact area of the region bounded by C, the x-axis and the line x = 8, and verify your answer by first converting C to Cartesian form.
Worked answer
x = 8 gives t = 2, and dx/dt = 3t2, so the area is ∫02 y (dx/dt) dt = ∫02 3t2 × 3t2 dt = ∫02 9t4 dt = [9t5/5] from 0 to 2 = 288/5. For the check, t = x1/3 gives y = 3x2/3, and ∫08 3x2/3 dx = [(9/5)x5/3] from 0 to 8 = 288/5. B1 for t = 2, M1 A1 for the integral of y times dx/dt, A1 for 288/5, M1 A1 for the Cartesian check. The mark most often dropped is the change of limits. x = 8 has to become t = 2 before the integral in t is written down, and candidates who carry 0 and 8 into a t-integral lose everything after the first line.Write the limit, as δx → 0, of the sum Σ x2 δx taken over strips from x = 1 to x = 3 as an integral, and evaluate it.
Worked answer
The limit is ∫13 x2 dx, and that evaluates to [x3/3] from 1 to 3 = 9 − 1/3 = 26/3. B1 for the integral with the right limits, M1 for integrating, A1 for 26/3. Each term f(x) δx is one strip's area, Σ adds the strips, and the limit turns the sum into the stretched S with dx sitting where δx used to.The curve C has parametric equations x = 3 cos t, y = sin t, 0 ≤ t ≤ π/2. Show that the area of the region bounded by C and the coordinate axes is 3π/4.
Worked answer
dx/dt = −3 sin t, and as t runs from 0 to π/2 the point travels leftwards from (3, 0) to (0, 1). Sweeping the other way, the area is ∫π/20 y (dx/dt) dt = 3∫0π/2 sin2 t dt. Now use sin2 t = ½ − ½ cos 2t, giving 3[t/2 − (sin 2t)/4] from 0 to π/2 = 3(π/4 − 0) = 3π/4 as required. M1 for the integral of y times dx/dt, A1 for handling the reversed limits, M1 A1 for the identity and the integration, A1 for the printed result. The identity is unavoidable, since sin2 t cannot be integrated as it stands. Leave a stray minus sign there and the printed answer can only be reached by a fudge, which shows in the working.The curve C has parametric equations x = 2 sin t, y = 3 sin 2t, 0 ≤ t ≤ π/2. Show that the area of the region between C and the x-axis is exactly 4.
Worked answer
dx/dt = 2 cos t, and x increases from 0 to 2 as t runs from 0 to π/2, so the area is ∫0π/2 3 sin 2t × 2 cos t dt. Replace sin 2t by 2 sin t cos t and the integrand becomes 12 sin t cos2 t. That is a reverse chain, since the derivative of cos3 t is −3 sin t cos2 t, so the antiderivative is −4 cos3 t. Evaluating from 0 to π/2 gives 0 − (−4) = 4. M1 for dx/dt, M1 A1 for the integral in t with the correct limits, M1 for replacing sin 2t by 2 sin t cos t, A1 for the antiderivative, A1 for the printed result. Most attempts stall trying to integrate 6 sin 2t cos t as it stands. The double angle has to be unpicked before any integration can start.
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