Questions on integration as antidifferentiation

Question 1

Find ∫(3x² − 2x + 5) dx.

Accepted Answer

Reverse the power rule term by term to get x³ − x² + 5x + c. M1 for reversing the power rule, A1 for the full answer including the + c.

Question 2

Explain why every indefinite integral carries a + c.

Accepted Answer

Differentiation destroys constants. Both x³ + 7 and x³ − 2 differentiate to 3x², so running the machine backwards recovers a whole family of curves, one for each constant. The + c holds the family's place until extra information picks a member. B1 for differentiation destroying constants, B1 for the family of curves the + c stands for.

Question 3

Find ∫(x + 2)(x − 5) dx.

Accepted Answer

The reversed power rule applies to powers of x, not to a product of brackets, so multiply out first: (x + 2)(x − 5) = x² − 3x − 10. M1 for the expansion. Integrating term by term gives x³/3 − 3x²/2 − 10x + c. A1 for the answer including the + c.

Question 4

Two curves have the same gradient function, dy/dx = 2x − 5. One passes through (0, 3) and the other through (0, 8). Find the equation of each curve, and state the vertical distance between them at x = 4.

Accepted Answer

Integrating gives y = x² − 5x + c, one curve for each value of c. M1 for the integration with its constant. Each point is on the y-axis, so it fixes c at once: the curves are y = x² − 5x + 3 and y = x² − 5x + 8. A1 for both equations. Subtracting, the second is always 5 above the first, whatever x is, so at x = 4 they are 5 apart, where their heights are −1 and 4. B1 for 5.

Question 5

Find ∫(6x⁵ − 3/x²) dx.

Accepted Answer

Rewrite the fraction as −3x⁻². The integral is then x⁶ + 3x⁻¹ + c, or x⁶ + 3/x + c. M1 for rewriting as a negative power, A1 for the first term, A1 for the second. Raising −2 by one gives −1, and dividing by −1 flips the sign. The double negative decides the sign, and an answer of x⁶ − 3/x + c has missed it.

Question 6

Find ∫(2√x + 1/√x) dx.

Accepted Answer

Write the integrand as 2x^1/2 + x^−1/2. Raising each index by one and dividing gives (4/3)x^3/2 + 2x^1/2 + c. M1 for writing both terms as powers of x, A1 for the first term, A1 for the second. Dividing by 3/2 means multiplying by 2/3, and dividing by 1/2 means doubling.

Question 7

A curve passes through the point (2, 5) and satisfies dy/dx = 3x² − 2. Find an equation for the curve.

Accepted Answer

Integrating gives y = x³ − 2x + c, a family of curves. The point picks the member. Substituting x = 2 and y = 5 gives 8 − 4 + c = 5, so c = 1 and y = x³ − 2x + 1. M1 for integrating, A1 for the family with its constant, M1 for substituting the point, A1 for the equation of the curve. Use the point to find c, and substitute into y rather than into dy/dx.

Question 8

A curve passes through (1, 5) and has gradient function dy/dx = 3x² − 4. A student writes 5 = 3(1)² − 4 + c and concludes that c = 6. Explain the error, and find the correct equation of the curve.

Accepted Answer

The point has been substituted into the gradient function rather than into the integrated equation. What the student's line actually says is that the gradient at x = 1 is 5, and it is not: 3(1)² − 4 = −1. B1 for the point going into dy/dx instead of into y. Integrate first: y = x³ − 4x + c. M1 for integrating, A1 for the family with its constant. Now the point belongs: 5 = 1 − 4 + c, so c = 8 and the curve is y = x³ − 4x + 8. A1 for c = 8 with the equation. The check is that (1, 5) satisfies the final equation: 1 − 4 + 8 = 5.

Question 9

A curve has gradient function dy/dx = 6x² + k, where k is a constant, and passes through the points (0, 1) and (2, 15). Find the value of k and the equation of the curve.

Accepted Answer

Integrate, carrying k through as the constant it is: y = 2x³ + kx + c. M1 for integrating, A1 for 2x³ + kx + c with two unknown constants. Two points are given because there are two unknowns. The first, (0, 1), sends every term but the constant to zero: 1 = 0 + 0 + c, so c = 1. M1 for using (0, 1) to get c = 1. The second gives 15 = 16 + 2k + 1, so 2k = −2. M1 for substituting (2, 15). Hence k = −1 and y = 2x³ − x + 1. A1 for both.

Question 10

The surface area of a sphere of radius r cm is 4πr² cm², and its volume V cm³ satisfies dV/dr = 4πr². Given that V = 0 when r = 0, find V in terms of r, and check that your answer has the dimensions of a volume.

Accepted Answer

Integrate with respect to r, treating 4π as a constant multiplier: V = 4π × r³/3 + c. M1 for integrating, A1 for (4/3)πr³ + c. A sphere of zero radius holds nothing, so V = 0 at r = 0 gives c = 0 and V = (4/3)πr³. A1 for the volume with the constant found. The dimensions check: r is a length in cm, so r³ is cm³, and π and 4/3 are pure numbers, so V comes out in cm³. B1 for the dimension check. For r = 3 the formula gives about 113 cm³.

Question 11

Find ∫(4x⁵ − 3x³ + 6)/(2x³) dx.

Accepted Answer

No rule integrates a quotient, so divide each term of the numerator by 2x³ before anything else: 4x⁵/(2x³) = 2x², 3x³/(2x³) = 3/2, and 6/(2x³) = 3x⁻³. M1 for dividing term by term, A1 for 2x² − 3/2 + 3x⁻³. Now reverse the power rule on each: 2x³/3, then −3x/2, and for the last term the index rises from −3 to −2 and divides, giving 3x⁻²/(−2). M1 for integrating all three powers. So the integral is 2x³/3 − 3x/2 − 3/(2x²) + c. A1 for the answer.

Question 12

Show that (x² + 1)³/6 + c is not an integral of (x² + 1)².

Accepted Answer

Differentiate the proposal. The chain rule gives 3(x² + 1)² × 2x ÷ 6 = x(x² + 1)², which carries an unwanted factor of x. The reversed power rule applies to powers of x, not to powers of a bracket, so the bracket has to be expanded first. Doing so gives x⁴ + 2x² + 1, whose integral is x⁵/5 + 2x³/3 + x + c. M1 for differentiating the proposal, A1 for the unwanted factor of x, B1 for the correct integral.

Question 13

Given that d²y/dx² = 6x − 4, that dy/dx = 5 when x = 1, and that y = 4 when x = 2, find y in terms of x.

Accepted Answer

Integrate once to get dy/dx = 3x² − 4x + c. The gradient condition gives 3 − 4 + c = 5, so c = 6. Integrate again to get y = x³ − 2x² + 6x + d. The point condition gives 8 − 8 + 12 + d = 4, so d = −8 and y = x³ − 2x² + 6x − 8. M1 for the first integration, A1 for dy/dx with its constant, M1 for using the gradient condition, M1 for the second integration, A1 for y with its second constant, A1 for the final equation. The fatal error is carrying a single c through both integrations, which quietly discards the cx term and makes the final answer unrecoverable.

Question 14

F and G are both antiderivatives of the same function f on an interval, so that F′(x) = G′(x) at every point of that interval. By considering H(x) = F(x) − G(x), prove that F and G differ by a constant, and explain what this proves about writing ∫f(x) dx = F(x) + c. Give two antiderivatives of 4x³ that differ by 8.

Accepted Answer

Differentiate H. Since differentiation is linear, H′(x) = F′(x) − G′(x), and the two derivatives are equal at every point of the interval, so H′(x) = 0 throughout it. M1 for H′ = F′ − G′ = 0. A function whose gradient is zero at every point of an interval has a horizontal graph there, so it takes one value throughout: H(x) = k for some constant k. A1 for zero gradient across an interval meaning constant. Hence F(x) = G(x) + k: any two antiderivatives of f differ by a constant and by nothing else. A1 for the conclusion. That is what makes F(x) + c the complete answer rather than one answer among others. Every antiderivative of f is F(x) + c for some c, so the notation names them all. B1 for what the + c claims. For an example, x⁴ + 5 and x⁴ − 3 both differentiate to 4x³ and differ by 8 at every x. B1 for a valid pair. The argument needs an interval: on a domain in two separate pieces the constant can be different on each, which is why the interval is stated in the setting-up.

Question 15

A tap fills a barrel. The rate at which the volume V litres of water in the barrel increases is modelled by dV/dt = 12 − 0.6t litres per minute, where t is the time in minutes since the tap was opened, and the barrel is empty when t = 0. Find V in terms of t, find the volume in the barrel after 20 minutes, and explain why the model must not be used for t > 20.

Accepted Answer

The right-hand side contains only t, so no separation is needed: this is an antiderivative with respect to t. Integrating, V = 12t − 0.3t² + c. M1 for integrating, A1 for 12t − 0.3t² + c. Empty at t = 0 gives 0 = 0 + c, so c = 0 and V = 12t − 0.3t². A1 for c = 0. At t = 20, V = 240 − 0.3 × 400 = 240 − 120. M1 for substituting t = 20. So the barrel holds 120 litres: litres per minute multiplied by minutes are litres. A1 for 120 litres with its units. The modelled rate is 12 − 0.6t, which is zero exactly at t = 20 and negative after it, so beyond 20 minutes the formula has the volume decreasing and the tap taking water out of the barrel. B1 for the rate changing sign at t = 20 as the reason.