Practise › Questions › Areas, parametric curves and the limit of a sum
Areas, parametric curves and the limit of a sum questions
Three extensions to finding areas. The region between two curves is one integral of top minus bottom, a curve given parametrically integrates without converting back to Cartesian form, and underneath both sits the definition of the definite integral as the limit of a sum of thinner and thinner strips.
7 original questions · 32 marks · the areas, parametric curves and the limit of a sum notes · Integration
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise areas, parametric curves and the limit of a sum one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.