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Integration as antidifferentiation questions
Integration reverses differentiation. It asks which function had this gradient, and answers with the power rule run backwards. An indefinite integral includes an arbitrary constant because differentiation removes additive constants, and only extra information can recover the one that was there.
7 original questions · 23 marks · the integration as antidifferentiation 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 ∫(3x2 − 2x + 5) dx.
Worked 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. Differentiating that answer lands back on the integrand, a free check worth the ten seconds. Omitting the + c costs a mark on almost every indefinite integral in the paper.Explain why every indefinite integral carries a + c.
Worked 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.Find ∫(6x5 − 3/x2) dx.
Worked 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. That double negative is where the marks are won and lost, and an answer of x6 − 3/x + c shows it was missed.Find ∫(2√x + 1/√x) dx.
Worked answer
Write the integrand as 2x1/2 + x−1/2. Raising each index by one and dividing gives (4/3)x3/2 + 2x1/2 + 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. Most of the work here is fraction arithmetic, and most of the lost marks are too.A curve passes through the point (2, 5) and satisfies dy/dx = 3x2 − 2. Find an equation for the curve.
Worked 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. Stopping at the family, or substituting into dy/dx instead of into y, loses the last two marks.Show that (x2 + 1)3/6 + c is not an integral of (x2 + 1)2.
Worked 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.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.
Worked 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. Two integrations mean two constants, and each condition pins exactly one of them. The fatal error is carrying a single c through both integrations, which quietly discards the cx term and makes the final answer unrecoverable. Using the gradient condition first keeps the arithmetic clean, though working the other way round and solving for both constants together is equally valid.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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