Questions on areas, parametric curves and the limit of a sum

Question 1

Find the area of the region bounded by the curve y = 4 − x², the x-axis and the y-axis, in the first quadrant.

Accepted 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.

Question 2

Find the area enclosed between the line y = x and the curve y = x².

Accepted 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. Choose top minus bottom for geometric area: the other order gives −1/6, and an area is not negative.

Question 3

Express ∫₁⁴ (2x + 3) dx as the limit of a sum, and state what a single term of that sum measures.

Accepted Answer

Cut the interval from x = 1 to x = 4, three units wide, into n strips of width δx = 3/n. The strip standing at x has height 2x + 3, so the strips together have total area Σ (2x + 3) δx, taken over strips from x = 1 to x = 4, and the integral is what that total tends to as the strips are made thinner: ∫₁⁴ (2x + 3) dx = the limit, as δx → 0, of Σ (2x + 3) δx. B1 for the limit with the right integrand and range. A single term (2x + 3) δx is the area of one strip, a rectangle of height 2x + 3 and width δx. B1 for the strip area. The integral sign is a stretched S standing for the sum, and dx is what δx becomes in the limit.

Question 4

The curve with equation y = x(4 − x) meets the line with equation y = 3 at two points. Find the x-coordinates of those points, and hence find the area of the finite region enclosed between the curve and the line.

Accepted Answer

Equate the two: x(4 − x) = 3, so 4x − x² = 3 and x² − 4x + 3 = 0. M1 for equating and rearranging to a quadratic equal to zero. Factorising, (x − 1)(x − 3) = 0, so the crossings are at x = 1 and x = 3. A1 for both values, which are the limits of the region. One test value settles the order: at x = 2 the curve is at 2 × 2 = 4 and the line is at 3, so the curve lies above the line for 1 < x < 3 and the height of a strip is curve minus line, x(4 − x) − 3 = −x² + 4x − 3. M1 for integrating that difference between those limits. Then ∫₁³ (−x² + 4x − 3) dx = [−x³/3 + 2x² − 3x] from 1 to 3 = (−9 + 18 − 9) − (−1/3 + 2 − 3) = 0 − (−4/3) = 4/3. A1 for 4/3. The lower boundary is horizontal, so the answer can be checked without a second integral: the area under the curve from 1 to 3 is [2x² − x³/3] from 1 to 3 = 9 − 5/3 = 22/3, and the rectangle under the line has area 2 × 3 = 6, leaving 22/3 − 6 = 4/3 again.

Question 5

Find the area enclosed between the curves y = 9 − x² and y = x² + 1.

Accepted 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.

Question 6

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.

Accepted 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 = x^1/3 gives y = 3x^2/3, and ∫₀⁸ 3x^2/3 dx = [(9/5)x^5/3] 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. Because the integral is with respect to t, convert the boundary x = 8 to t = 2 before setting the limits.

Question 7

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.

Accepted 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.

Question 8

Find the total area of the two finite regions enclosed between the curve y = x³ and the line y = x.

Accepted Answer

Solve x³ = x for the crossings: x³ − x = x(x − 1)(x + 1) = 0, so the graphs meet at x = −1, x = 0 and x = 1. M1 for solving, A1 for all three values. The two graphs change places at the origin. At x = ½ the line is at 0.5 and the curve at 0.125, so the line is the upper graph on the right; at x = −½ the curve is at −0.125 and the line at −0.5, so the curve is the upper graph on the left. The two regions therefore have to be integrated separately. B1 for splitting at x = 0 with the reason. On [0, 1], ∫(x − x³) dx = [x²/2 − x⁴/4] from 0 to 1 = 1/2 − 1/4 = 1/4. On [−1, 0], ∫(x³ − x) dx = [x⁴/4 − x²/2] from −1 to 0 = 0 − (1/4 − 1/2) = 1/4. M1 for integrating an upper graph minus a lower one over one of the regions. The total area is 1/2. A1 for both quarters and the total. Integrating x − x³ straight through from −1 to 1 returns 0, because the two signed pieces cancel, and the symmetry that makes them cancel is the same symmetry that makes the two areas equal.

Question 9

The finite region R is bounded by the curve y = √x, the line y = x − 2 and the x-axis. Find the exact area of R.

Accepted Answer

Where the curve meets the line, √x = x − 2. Squaring gives x = x² − 4x + 4, so x² − 5x + 4 = 0 and (x − 1)(x − 4) = 0. M1 for squaring and solving. At x = 1 the curve is at 1 and the line at −1, so squaring has manufactured a root that is not a crossing; at x = 4 both are 2. The graphs meet only at (4, 2). A1 for x = 4 with x = 1 tested and rejected. The line cuts the x-axis at (2, 0), so R is the region under the curve from x = 0 to x = 4 with the triangle below the line taken away. Under the curve, ∫₀⁴ x^1/2 dx = [(2/3)x^3/2] from 0 to 4 = 16/3. M1 for the integral, A1 for 16/3. The triangle has base 4 − 2 = 2 and height 2, so its area is 2, and R has area 16/3 − 2 = 10/3. A1 for 10/3.

Question 10

The curve C has parametric equations x = t², y = t(4 − t), 0 ≤ t ≤ 4. Find the exact area of the finite region enclosed between C and the x-axis.

Accepted Answer

No value of x is given, so the limits have to come from the curve itself. C meets the x-axis where y = 0, that is where t(4 − t) = 0, so t = 0 and t = 4, and y ≥ 0 in between. B1 for t = 0 and t = 4 read off y = 0. Then dx/dt = 2t, and x increases from 0 to 16 as t runs from 0 to 4, so the area is ∫₀⁴ y (dx/dt) dt = ∫₀⁴ t(4 − t) × 2t dt. M1 for y times dx/dt with t-limits. The integrand multiplies out to 8t² − 2t³. A1 for the integrand in t. Integrating, [8t³/3 − t⁴/2] from 0 to 4 = 512/3 − 128 = 128/3. M1 for integrating and substituting, A1 for 128/3. The parameter does the work of finding the limits here: solving y = 0 in t is one line, while converting to Cartesian form first means dealing with y = √x(4 − √x) and a root in every term.

Question 11

The curve with equation y = 6/x and the line with equation y = 7 − x enclose a finite region. Find the exact area of that region.

Accepted Answer

Equate: 6/x = 7 − x. Multiplying by x gives 6 = 7x − x², so x² − 7x + 6 = 0 and (x − 1)(x − 6) = 0. M1 for equating and clearing the fraction, A1 for x = 1 and x = 6. Between them the line is the upper graph: at x = 2 the line is at 5 and the curve at 3. So the area is ∫₁⁶ (7 − x − 6/x) dx = [7x − x²/2 − 6 ln x] from 1 to 6. M1 for integrating upper minus lower, A1 for the antiderivative. No modulus is needed in the logarithm, since x runs from 1 to 6 and stays positive. At x = 6 the bracket is 42 − 18 − 6 ln 6 = 24 − 6 ln 6, and at x = 1 it is 7 − 1/2 − 0 = 13/2, so the area is 35/2 − 6 ln 6, about 6.75. A1 for the exact value. The question asks for an exact area, so the logarithm stays in the answer.

Question 12

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.

Accepted 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∫₀^π/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. Leave a stray minus sign in the limits and the printed answer can only be reached by a fudge, which shows in the working.

Question 13

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.

Accepted 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 ∫₀^π/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.

Question 14

A region is bounded above by the curve y = g(x), below by the curve y = f(x), and at its sides by the lines x = a and x = b, where a < b and g(x) > f(x) throughout. By considering a thin strip of width δx, explain why the area of the region is ∫_a^b [g(x) − f(x)] dx. Explain why the same expression gives the area when the whole region lies below the x-axis, and state what a negative value of the integral would show.

Accepted Answer

Take the strip standing at x, running from the lower curve up to the upper one. Its width is δx and its height is the vertical gap g(x) − f(x), so its area is very nearly [g(x) − f(x)] δx, a rectangle. B1 for the strip's area as height times width. Adding the strips across the region gives Σ [g(x) − f(x)] δx, and as δx → 0 the rectangles' error at the curved top and bottom of each strip vanishes, so the sum tends to ∫_a^b [g(x) − f(x)] dx. B1 for the sum and the limit. Nothing in that argument used the position of the x-axis. The height of the strip is a difference, and moving the whole picture up by k replaces g by g + k and f by f + k, leaving g − f unchanged. B1 for the height being a difference, so the axis does not enter. If the region lies wholly below the axis, both g(x) and f(x) are negative, but g(x) − f(x) is still positive, so the integral is still the positive area. B1 for the below-the-axis case. A negative value means g(x) − f(x) is negative over the range, that is, the curve called g was not the upper one: the subtraction has been taken the wrong way round, and the area is the size of the answer. B1 for that reading.

Question 15

The curve C has parametric equations x = t², y = t³ − 3t. Show that C is symmetrical about the x-axis, and find the exact area of the loop that C encloses.

Accepted Answer

Replacing t by −t leaves x = t² unchanged and turns y = t³ − 3t into −(t³ − 3t), so the points at parameters t and −t have the same x and opposite y: each is the reflection of the other in the x-axis, and C is symmetrical about that axis. B1 for the symmetry, argued from the parameters. C crosses the x-axis where y = 0, that is where t(t² − 3) = 0, so t = 0 at the origin and t = ±√3, both giving the single point (3, 0). The curve leaves (3, 0) at t = −√3, passes through the origin at t = 0 and returns to (3, 0) at t = √3, so that stretch of parameter is the loop. B1 for the loop running from t = −√3 to t = √3 through the origin. By the symmetry it is enough to work with the half for 0 ≤ t ≤ √3, where y ≤ 0 and x increases from 0 to 3. With dx/dt = 2t, ∫₀^√3 y (dx/dt) dt = ∫₀^√3 (t³ − 3t) × 2t dt. M1 for y times dx/dt over the half loop. The integrand is 2t⁴ − 6t². A1 for the integrand in t. Integrating, [2t⁵/5 − 2t³] from 0 to √3 = 18√3/5 − 6√3 = −12√3/5, using (√3)⁵ = 9√3. M1 for integrating and substituting. The value is negative because that arc lies below the axis, so that half of the loop has area 12√3/5 and the whole loop has area 24√3/5, about 8.31. A1 for the exact area. Working with the half and doubling keeps the sign easy to read. Taken in one sweep from t = −√3 to t = √3 the two halves do not cancel: y and dx/dt change sign together, so the integrand 2t⁴ − 6t² is even and the integral is −24√3/5, the whole loop with a sign set by the direction of travel.