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Integration as antidifferentiation

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.

Year 12-13EDEXCEL 9MA0 8.1, 8.2

Builds on Differentiating powers of x.

IN THIS TOPIC

  • Integrate powers of x by reversing the power rule, with the constant of integration.
  • Rewrite expressions into powers before integrating, and check answers by differentiating.
  • Recover a curve from its gradient function and one known point.

COMMON MISCONCEPTION

∫xn dx = xn+1/(n + 1), and that is the whole answer.

Reversing differentiation

Integration undoes differentiation. Made precise, that sentence is the Fundamental Theorem of Calculus, and it turns every derivative fact you know into a fact about integrals. Reverse the power rule by adding one to the exponent and then dividing by the new exponent.

∫xndx=xn+1n+1+c(n≠−1)\text{∫} x^{n} \, dx = \frac{x^{n+1}}{n + 1} + c \, (n ≠ −1)NOT IN THE BOOKLET — LEARN IT

Two footnotes come attached. The + c is compulsory, because differentiation removes additive constants and the reverse process cannot determine which one was there. And n = −1 is excluded, since it would require a division by zero. Its integral turns up in Year 13.

Differentiation sends x cubed plus any constant to the same 3 x squared, so integrating can only recover x cubed plus an unknown constant cx³ − 2x³x³ + 53x²differentiateintegratex³ + c
FIG. 1Why the c is not optional: x³ − 2, x³ and x³ + 5 all differentiate to 3x². Running backwards from 3x² can only give x³ + c.

WORKED EXAMPLE

Termwise, with the constant

Find ∫(6x2 − 4x + 3) dx.

Integrate each term by the reversed power rule. 6x2 becomes 2x3, −4x becomes −2x2, and 3 becomes 3x.

So the integral is 2x3 − 2x2 + 3x + c.

Differentiating the answer is a free check, and it lands back on 6x2 − 4x + 3 on the nose. Every integration in this course can be checked that way, and the check takes seconds.

GUIDED PRACTICE

Rewrite first, as ever

Find ∫(½x2 − 3/√x) dx, before opening the working.

Show the working

Rewrite the second term as a power: −3x−1/2.

Integrating termwise gives x3/6 − 3 × 2x1/2 + c, that is x3/6 − 6√x + c.

The −½ exponent went up to +½, and dividing by ½ doubled the coefficient. Fractional exponents make sign and arithmetic slips easy, and the differentiate-back check catches all of them.

One point pins the constant

An indefinite integral is not one curve but a whole family, one member for each value of c, every one of them sharing the same gradient at every x. To single out one member a question gives you a point the curve passes through. Substitute it and c is determined.

The family of curves with gradient function 2x: parabolas x squared plus c, with the member through 2 comma 3 fixing c equal to minus 1(2, 3)one point picks one curve
FIG. 2The family with gradient function 2x, one parabola per c. The marked point (2, 3) belongs to exactly one member, and finding c is one substitution.

INDEPENDENT PRACTICE

From gradient to curve

A curve has gradient function dy/dx = 3x2 − 8x and passes through (2, 3). Find its equation.

Show the working

Integrate: y = x3 − 4x2 + c.

Substitute the point: 3 = 8 − 16 + c, so c = 11.

The curve is y = x3 − 4x2 + 11, and no other member of the family goes through (2, 3).

Integrate first, substitute second. Substituting into the gradient function instead is the classic wrong turn, and it produces a gradient, never a c.

ASSESSMENT FOCUS

  • Add one to the exponent, divide by the new exponent, and put + c on every indefinite integral. A missing c costs a mark every time.
  • Rewrite roots, reciprocals and quotients as powers before integrating, exactly as you did for differentiation.
  • Check by differentiating back. The check is silent, fast, and catches almost every slip this topic produces.
  • Given dy/dx and a point, integrate first and substitute second, then state the full equation as your answer.

CHECK YOURSELF

Find ∫(4x3 + 2/x3) dx, and verify your answer by differentiation.

Show a hint

2/x³ is 2x⁻³; the exponent climbs to −2.

Show the answer

Rewriting and integrating termwise, ∫(4x3 + 2x−3) dx = x4 − x−2 + c, that is x4 − 1/x2 + c.

Differentiating back gives 4x3 + 2x−3, the original integrand, so the answer stands.

The negative exponent rose from −3 to −2, and dividing by −2 flipped the sign. Both steps are where the marks are awarded.

Raise the exponent by one, divide by it, and never leave without the + c.

A gradient function names a family; one known point picks the member.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 31 KB · print edition · white background

Question workbook

PDF · 31 KB · screen edition · navy

Answer book

PDF · 36 KB · fully worked · navy
7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the integration as antidifferentiation questions page.

CHECK YOUR PROGRESS

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  • Integrate powers of x by reversing the power rule, with the constant of integration.
  • Rewrite expressions into powers before integrating, and check answers by differentiating.
  • Recover a curve from its gradient function and one known point.

Open the full revision checklist to see every objective in the course in one place.