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Integration as antidifferentiation

Integration reverses differentiation: it finds the functions with a given derivative, using the power rule in reverse. An indefinite integral includes an arbitrary constant, because differentiation removes additive constants; a further condition is needed to determine it.

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 the conditions that pin its constants.
  • Explain why the constant of integration is required, and why two antiderivatives of the same function differ only by a constant.

COMMON MISCONCEPTION

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

Two conditions come with that formula. The + c is required, because differentiation removes additive constants: x3 − 2, x3 and x3 + 5 all differentiate to 3x2, so running backwards can only give x3 + c. And n = −1 is excluded, since the formula would divide by zero.

Reversing differentiation

Integration reverses differentiation, a statement made precise by the Fundamental Theorem of Calculus, so every derivative result gives a corresponding integral. 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 conditions come with it. The + c is required, because differentiation removes additive constants, so the reverse process determines the function only up to a constant. And n = −1 is excluded, since the formula would divide by zero; ∫(1/x) dx = ln|x| + c is treated in Integrating standard functions.

Three grey boxes, x cubed minus 2, x cubed and x cubed plus 5, all feeding one cyan box labelled 3 x squared under differentiation, with a single amber arrow running back to x cubed plus c: the reverse journey cannot know which constant was lost, so it must carry a 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 returns 6x2 − 4x + 3, which checks the integration. Every integral can be checked this way.

GUIDED PRACTICE

Rewrite as powers first

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 exponent −½ rose to +½, and dividing by ½ doubled the coefficient. Differentiating the answer checks both steps.

One point pins the constant

An indefinite integral defines a family of curves, one for each value of c, all with the same gradient at each x. A point on the required curve determines c: substitute its coordinates into the integrated equation and solve.

A stack of computed parabolas differing only by their constant c, all sharing the gradient function 2x, with the amber member through the marked point 2 comma 3 singled out: knowing one point on the curve is what pins the constant down.
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, the only member of the family through (2, 3).

Integrate first and substitute second. Substituting the point into the gradient function gives the gradient at that point, not the value of c.

ASSESSMENT FOCUS

  • Add one to the exponent, divide by the new exponent, and include + c on every indefinite integral: without it the answer names one antiderivative rather than all of them.
  • Rewrite roots, reciprocals and quotients as powers before integrating, as for differentiation.
  • Check by differentiating the answer back to the integrand.
  • 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 increases 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 exponent rose from −3 to −2, and dividing by −2 changed the sign of the term. Show both steps.

Raise the exponent by one, divide by the new exponent, and include the + c.

A gradient function determines a family of curves; one known point determines which 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

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Question workbook

PDF · 76 KB · screen edition · navy

Answer book

PDF · 85 KB · fully worked · navy
15 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.

USED LATER IN

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 the conditions that pin its constants.
  • Explain why the constant of integration is required, and why two antiderivatives of the same function differ only by a constant.

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