Maths › Integration › Integration as antidifferentiation

Integration as antidifferentiation

Run differentiation backwards and a new subject falls out. Integration asks which function had this gradient, answers with the power rule reversed, and carries an honest confession in every answer: a constant was lost on the way down, and only extra information can bring it back.

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.

WHAT YOU PROBABLY THINK

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

Running the machine backwards

Integration undoes differentiation; that statement, made precise, is the Fundamental Theorem of Calculus, and it makes every derivative fact a fact about integrals too. Reversing the power rule, add one to the exponent, then divide by the new exponent,

∫ xn dx = xn+1n + 1 + c (n ≠ −1)NOT IN THE BOOKLET — LEARN IT

with two footnotes the opening lie ignores. The + c is compulsory, because differentiation destroys constants and the road back cannot know which one was lost. And n = −1 is excluded, since it would demand division by zero; its integral arrives with Year 13.

Differentiation sends x cubed, x cubed plus 5 and x cubed minus 2 all to the same gradient function 3 x squared, so running the machine backwards 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 honestly say 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 can be checked this way, and the habit costs seconds.

YOUR TURN

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 the division by ½ doubled the coefficient. Fractional exponents make sign and arithmetic slips easy, and the differentiate-back check catches them all.

One point pins the constant

An indefinite integral is a whole family of curves, one for each value of c, all sharing the same gradient everywhere. To single out one member, a question supplies a point the curve passes through, and substituting it turns c from unknown to known.

The family of curves with gradient function 2x: parabolas x squared plus c for several values of c, with the single member through the point 2 comma 3 picked out, which fixes 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.

TRY IT UNSEEN

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 passes 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.

THE EXAM BIT

  • Add one to the exponent, divide by the new exponent, and write + c on every indefinite integral; a missing c drops a mark every time it happens.
  • Rewrite roots, reciprocals and quotients as powers before integrating, exactly as 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, then substitute the point to find c, and state the full equation as the answer.
  • n = −1 is excluded from the rule; if 1/x appears, the question belongs to a later lesson, not to a forced x0/0.

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 moves are where the marks in this question live.

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.

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.

No animated video for this topic yet; these notes stand alone.