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Definite integrals and areas questions
Put limits on an integral and the constant cancels, the answer becomes a number, and the number means something: the area between curve and axis. Regions below the axis contribute negatively, so finding an area means establishing where the curve crosses the axis before integrating across it.
6 original questions · 22 marks · the definite integrals and areas 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.
Evaluate ∫02 x3 dx.
Worked answer
Antidifferentiate inside square brackets, [x4/4], then substitute. 16/4 − 0 = 4. M1 for the antiderivative in brackets, A1 for 4. Upper limit first, lower limit subtracted. No + c appears, since the same constant would sit in both substitutions and cancel.Evaluate ∫14 (2x + 1) dx.
Worked answer
[x2 + x] from 1 to 4 = 20 − 2 = 18. M1 for the antiderivative, A1 for 18. Geometry checks it. The region is a trapezium with parallel sides 3 and 9 and width 3, and ½(3 + 9) × 3 = 18. Marks go missing here when the limits are substituted the wrong way round.Find the area of the finite region enclosed between the curve y = 6x − x2 and the x-axis.
Worked answer
Find the roots before anything else. 6x − x2 = x(6 − x) = 0 at x = 0 and x = 6, and the arch lies above the axis between them. Area = ∫06 (6x − x2) dx = [3x2 − x3/3] from 0 to 6 = 108 − 72 = 36. M1 for solving for the roots, A1 for 0 and 6, M1 for the integration, A1 for 36. A candidate who invents limits rather than solving for them loses the first method mark.Find the total area enclosed between the curve y = x(x − 2) and the x-axis from x = 0 to x = 3.
Worked answer
The curve dips below the axis between its roots 0 and 2, then climbs above it. Split there. With F(x) = x3/3 − x2, the integral from 0 to 2 is −4/3, so that piece has area 4/3, and from 2 to 3 it is +4/3. Total area 8/3. B1 for splitting at the root, M1 for integrating a piece, A1 for −4/3, A1 for the second piece, A1 for the total 8/3. A single integral from 0 to 3 returns 0, because the signed pieces cancel. The split and the change of sign are the whole of this question, so an unsplit answer of zero has not addressed it.Show that ∫14 (x2 + 3)/√x dx = 92/5.
Worked answer
Divide through by √x first, giving x3/2 + 3x−1/2. Integrating term by term gives (2/5)x5/2 + 6x1/2. At x = 4 that is 64/5 + 12 = 124/5, and at x = 1 it is 2/5 + 6 = 32/5. The difference is 92/5, as required. M1 for dividing through by √x, A1 for the two powers, M1 for integrating, A1 for 92/5. No rule integrates a quotient directly, so the algebra has to happen before the integral sign acts. A show-that question also wants the exact fraction, never 18.4.The curve C has equation y = x2 − 2x and the line l has equation y = 3. Show that the finite region enclosed by C and l has area 32/3.
Worked answer
Solve x2 − 2x = 3 for the intersections. x2 − 2x − 3 = (x − 3)(x + 1) = 0, so C and l meet at x = −1 and x = 3. Between those values the line lies above the curve, so integrate line minus curve. ∫−13 (3 + 2x − x2) dx = [3x + x2 − x3/3] from −1 to 3 = 9 − (−5/3) = 32/3, as required. M1 for equating and solving, A1 for the limits −1 and 3, M1 for integrating line minus curve, A1 for the antiderivative, A1 for 32/3. Two slips dominate. Subtracting the wrong way round returns −32/3, and integrating the curve alone throws away the strip of area between the curve and the line. A sketch, or a single test value such as x = 0, settles which graph is on top and secures the method mark.
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