Questions on definite integrals and areas

Question 1

Evaluate ∫₀² x³ dx.

Accepted 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.

Question 2

Evaluate ∫₁⁴ (2x + 1) dx.

Accepted 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.

Question 3

Evaluate ∫₋₁² (3x² − 2x) dx.

Accepted Answer

Antidifferentiate inside square brackets, [x³ − x²], then substitute. At the upper limit that is 8 − 4 = 4; at the lower limit it is (−1)³ − (−1)² = −1 − 1 = −2. M1 for the antiderivative with both limits substituted. Subtracting, 4 − (−2) = 6. A1 for 6. Take both powers of the negative limit before subtracting: the cube keeps the sign and the square does not, and the minus in front of the lower value then turns −2 into +2.

Question 4

Evaluate ∫₁³ (2x + 5) dx + ∫₃⁶ (2x + 5) dx by first writing the two integrals as one, and state the property of definite integrals you have used.

Accepted Answer

The two intervals share the integrand and meet at x = 3, where one ends and the other begins, so between them they cover [1, 6] with nothing counted twice and nothing left out: the two integrals add to ∫₁⁶ (2x + 5) dx. B1 for combining them, naming the property that a definite integral adds over adjacent intervals meeting at a common limit. Then [x² + 5x] from 1 to 6 = (36 + 30) − (1 + 5) = 66 − 6. M1 for the antiderivative with both limits. The value is 60. A1 for 60. Working the two out separately gives 18 and 42, which also make 60, and that check costs one line. The property needs the upper limit of the first to be the lower limit of the second; intervals that overlap or leave a gap do not combine like this.

Question 5

Find the area of the finite region enclosed between the curve y = 6x − x² and the x-axis.

Accepted Answer

Find the intersections with the x-axis to determine the integration limits: 6x − x² = x(6 − x) = 0, so x = 0 and x = 6. 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.

Question 6

Find the total area enclosed between the curve y = x(x − 2) and the x-axis from x = 0 to x = 3.

Accepted 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.

Question 7

Show that ∫₁⁴ (x² + 3)/√x dx = 92/5.

Accepted Answer

Divide through by √x first, giving x^3/2 + 3x^−1/2. Integrating term by term gives (2/5)x^5/2 + 6x^1/2. 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.

Question 8

Using algebra, calculate the exact value of ∫₁⁴ (x − 3)²/√x dx.

Accepted Answer

Expand and divide through by √x before integrating anything: (x² − 6x + 9)/√x = x^3/2 − 6x^1/2 + 9x^−1/2. M1 for the expansion and division, A1 for the three powers. Integrating term by term gives [(2/5)x^5/2 − 4x^3/2 + 18x^1/2]. At x = 4 that is 64/5 − 32 + 36 = 84/5, and at x = 1 it is 2/5 − 4 + 18 = 72/5. M1 for integrating the powers and substituting both limits. The value is 84/5 − 72/5 = 12/5. A1 for the exact fraction. The question says exact, so 2.4 is not the required form.

Question 9

Given that ∫₀^k (2x + 1) dx = 12, where k > 0, find the value of k.

Accepted Answer

Integrate first and carry k through as the upper limit: [x² + x] from 0 to k = k² + k − 0. M1 for the antiderivative with k in place, A1 for k² + k. Setting that equal to 12 gives k² + k − 12 = 0, which factorises as (k + 4)(k − 3) = 0. M1 for forming and solving the quadratic. The roots are k = −4 and k = 3, and the condition k > 0 leaves k = 3. A1 for k = 3 with −4 rejected. Checking, [x² + x] from 0 to 3 = 9 + 3 = 12. The unknown is a limit rather than part of the integrand, so it plays no part in the integration and appears only when the limits are substituted.

Question 10

Evaluate ∫₀³ (x² − 4x) dx, state what the sign of your answer shows about the region between the curve and the x-axis for 0 ≤ x ≤ 3, and state the area of that region.

Accepted Answer

[x³/3 − 2x²] from 0 to 3 = (9 − 18) − 0. M1 for the antiderivative with both limits. The value is −9. A1 for −9. The curve is y = x(x − 4), so it meets the axis at x = 0 and x = 4 and at no point between them: across the whole of 0 < x < 3 it lies below the axis, and at x = 1, for instance, y = −3. The negative value records where the region is, not a negative size. B1 for the region lying below the axis throughout, with the two roots as the reason. The area is therefore 9. B1 for 9. Nothing had to be split here, because the only root in [0, 3] is the endpoint x = 0; an interval reaching past x = 4 would have picked up pieces of both signs and those would have had to be taken separately.

Question 11

The region R is bounded by the curve y = 8/x², the x-axis and the lines x = 1 and x = 4. Find the exact area of R, and find the value of a for which the line x = a divides R into two regions of equal area.

Accepted Answer

Write the integrand as a power, 8x⁻²: the index rises to −1 and divides by it, giving −8/x. M1 for the antiderivative as a negative power. The area is then [−8/x] from 1 to 4 = −2 − (−8) = 6. A1 for 6. Half of that is 3, and the left-hand piece runs from 1 to a: [−8/x] from 1 to a = 8 − 8/a. Setting 8 − 8/a = 3 gives 8/a = 5. M1 for equating the left-hand piece to half the area, A1 for 8/a = 5. So a = 8/5. A1 for a = 8/5, which lies between 1 and 4 as it must. It sits well to the left of the midpoint x = 2.5, because the curve is far taller near x = 1 than near x = 4, so the left-hand strip reaches half the area early.

Question 12

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.

Accepted 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.

Question 13

A student writes ∫_a^b f(x) dx = F(b) − F(a) + c, and separately claims that reversing the two limits leaves the value unchanged. Explain why no + c belongs in the value of a definite integral, and what reversing the limits actually does.

Accepted Answer

Whatever constant the antiderivative carries appears in both substitutions: the value is (F(b) + c) − (F(a) + c), and the two copies of c cancel in the subtraction, which is why square-bracket working never writes one. B1 for the cancellation. Reversing the limits swaps the roles in that subtraction, giving F(a) − F(b) = −(F(b) − F(a)), so the value does not survive: its sign flips, and so does the sign of everything computed from it downstream. B1 for the sign change. A region wholly above the axis integrated with reversed limits therefore comes out negative, and the honest response is to put the limits in increasing order and quote the area as a positive size, not to drop a stray minus sign at the end. B1 for the consequence for areas.

Question 14

A function f is odd if f(−x) = −f(x) for every x, so that its graph has rotational symmetry of order 2 about the origin. Explain why ∫_−a^a f(x) dx = 0 for every odd f and every a > 0, and write down the value of ∫₋₃³ (x⁵ − 2x³ + x) dx. Explain why the same argument reaches no such conclusion for ∫₋₃³ (x⁴ − x²) dx, and evaluate that integral.

Accepted Answer

Rotating the graph half a turn about the origin carries the part over [−a, 0] onto the part over [0, a], so the two regions have the same size. The same rotation exchanges above the axis for below, so the two signed contributions are equal in size and opposite in sign, and they cancel: the integral is 0. B1 for the cancellation argued from the symmetry. Every power in x⁵ − 2x³ + x is odd, so the function is odd and the value is 0. B1 for 0, with the function identified as odd. In x⁴ − x² both powers are even, so f(−x) = f(x). Here it is a reflection in the y-axis, not a rotation about the origin, that carries one half onto the other, and a reflection leaves the sign of the contribution alone. The two halves therefore agree in sign, nothing cancels, and the integral is twice one half. B1 for the even case giving twice rather than zero. So the value is 2∫₀³ (x⁴ − x²) dx = 2[x⁵/5 − x³/3] from 0 to 3. M1 for doubling the integral over [0, 3]. The bracket is 243/5 − 9 = 198/5. A1 for 198/5. The value is 396/5. A1 for 396/5. Integrating straight through from −3 to 3 gives the same number, so the symmetry is a shortcut and a check rather than the only route.

Question 15

The finite region R is bounded by the curve y = x³, the y-axis and the line y = 27. A student writes down ∫₀³ x³ dx as the area of R. Explain what that integral does measure, and find the area of R.

Accepted Answer

The curve meets y = 27 where x³ = 27, that is at x = 3, and it meets the y-axis at the origin, so R is the region lying above the curve and below the line between x = 0 and x = 3. The integral ∫₀³ x³ dx measures the region between the curve and the x-axis over the same range, which is the region on the other side of the curve. The two together fill the rectangle, and they are not the same region. B1 for what the integral measures. That rectangle runs from x = 0 to x = 3 and from y = 0 to y = 27, so its area is 3 × 27 = 81. M1 for the rectangle. The integral itself is [x⁴/4] from 0 to 3 = 81/4. A1 for 81/4. Taking the region below the curve away from the rectangle, M1 for the subtraction, R has area 81 − 81/4 = 243/4. A1 for 243/4. The check is that 243/4 is three times 81/4: a cubic rises slowly and then steeply, so most of the rectangle lies above it, and an answer smaller than 81/4 would be the wrong way round.

Question 16

The diagram shows the curve with equation y = 4x − x² and a straight line L drawn on the same grid. Use the diagram to write down the equation of L. Find the coordinates of the two points where L meets the curve, and hence find the exact area of the region enclosed between the curve and L.

Accepted Answer

L passes through the grid points (0, 0) and (3, 3), so its gradient is 3/3 = 1 and it cuts the y-axis at 0: its equation is y = x. B1 for the equation of L. Where the line meets the curve, 4x − x² = x, so 3x − x² = 0 and x(3 − x) = 0. M1 for equating the two and solving. That gives x = 0 and x = 3, and the line supplies the y values, so the points are (0, 0) and (3, 3). A1 for both points. Between x = 0 and x = 3 the curve lies above the line, so the area is the integral of the upper minus the lower: ∫₀³ (4x − x² − x) dx = ∫₀³ (3x − x²) dx. M1 for the difference of the two functions with limits 0 and 3. Antidifferentiating inside square brackets, [3x²/2 − x³/3] from 0 to 3 = (27/2 − 27/3) − 0 = 13.5 − 9. A1 for the antiderivative with both limits substituted. The area is 9/2. A1 for the exact value. Subtracting the two integrals separately gives the same thing, but doing the subtraction before integrating halves the work; and taking the curve alone would include the triangle under L that the region does not contain.