CA8–CA10 - Integration as the reverse of differentiation

CA8–CA10 - Integration as the reverse of differentiation

Differentiation finds a gradient from a curve; integration can work back from the gradient to the curve. CA8 gives the polynomial integration rule, CA9 explains why it reverses differentiation, and the indefinite-integral part of CA10 introduces the constant of integration. You will use one known point to select the right curve from a whole family.

Recovering a curve from its gradient

A curve has gradient 2x2x at every point and passes through (3,14)(3, 14). Is its equation y=x2y=x^2? That has the right gradient but the wrong height: at x=3x=3, it gives 99. How can we keep every gradient and still reach the given point?

You already know

The derivative of x2x^2 is 2x2x, and the derivative of a constant is zero. A vertical translation adds the same constant to every height without changing gradients.

The curves y=x2+7y=x^2+7, y=x2−4y=x^2-4 and y=x2y=x^2 all differentiate to 2x2x. The derivative has lost the constant. Reversing the process must restore that freedom rather than pretending the constant was definitely zero.

Indefinite integral

An indefinite integral represents a family of functions with the same derivative. If F′(x)=f(x)\mathrm{F}'(x)=\mathrm{f}(x), then ∫f(x) dx=F(x)+C\int\mathrm{f}(x)\,\mathrm{d}x=\mathrm{F}(x)+C, where CC is an arbitrary constant. The expression being integrated is the integrand, and dx\mathrm{d}x says that xx is the variable of integration.

The integral sign asks for the reverse operation: “What differentiates to this?” For example, ∫2x dx=x2+C\int2x\,\mathrm{d}x=x^2+C. Different choices of CC give vertically translated parabolas. An indefinite integral has no upper or lower limits; definite integrals, which have limits and produce a number, come next.

[DIAGRAM: asset_name: A family with the same gradient; asset_slug: ca34-integral-family; caption: All three parabolas have gradient 2x, but only one passes through 3 comma 14; description: Three parabolas y equals x squared, y equals x squared plus 5 and y equals x squared minus 4. The middle-height blue curve y equals x squared plus 5 passes through P at 3 comma 14. Matching small tangent segments at x equals 1 have gradient 2 on each curve.]
All three parabolas have gradient 2x, but only one passes through 3 comma 14

There is an exact reason that the missing information is just a constant for polynomials. Subtract two polynomials with identical derivatives. The difference has derivative zero. Any non-constant term would leave a non-zero derivative term, so the difference must be constant. Even complete gradient information cannot reveal a vertical position without an extra condition.

The inverse relationship and the constant of integration are not on the formulae sheet: learn them.

Key idea

Integration recovers a family of curves; one known point can fix the missing constant and identify one member.

Check what “arbitrary constant” means before learning the rule.

Integrating a power

To reverse the power rule, increase the power by one and divide by the new power.

For example, x3x^3 differentiates to 3x23x^2, so to obtain a derivative of just x2x^2 we need 13x3\frac13x^3. The division compensates for the multiplier that differentiation will produce. Integrating a constant works too: 7x7x differentiates to 77, so ∫7 dx=7x+C\int7\,\mathrm{d}x=7x+C.

Integrating a polynomial term

∫kxn dx=kn+1xn+1+C\int kx^n\,\mathrm{d}x=\frac{k}{n+1}x^{n+1}+C

Here kk is constant and nn is a positive integer or zero. In particular, ∫k dx=kx+C\int k\,\mathrm{d}x=kx+C. The constant CC can take any real value.
Not on the formulae sheet: learn it.

The check is built into the method: differentiate your answer and see whether the original integrand returns.

Worked example 1: a positive term

Question. Find ∫6x2 dx\int6x^2\,\mathrm{d}x.

Plan. Increase the power from 22 to 33, divide the coefficient by 33 and include the arbitrary constant.

∫6x2 dx=63x3+Cdivide by the new power=2x3+Csimplify the coefficient\begin{aligned} \int6x^2\,\mathrm{d}x&=\frac63x^3+C && \text{divide by the new power}\\ &=2x^3+C && \text{simplify the coefficient} \end{aligned}

Why divide by 33? Differentiating a cubic multiplies its coefficient by 33; division reverses that multiplication.

Answer. 2x3+C2x^3+C.

Check. ddx(2x3+C)=6x2\dfrac{\mathrm{d}}{\mathrm{d}x}(2x^3+C)=6x^2, exactly the integrand.

Change only the coefficient's sign. The new power is unchanged; the negative sign must survive.

Worked example 2: a negative term

Question. Find ∫−6x2 dx\int-6x^2\,\mathrm{d}x.

Plan. Use the same power increase, retaining the negative coefficient.

∫−6x2 dx=−63x3+Cincrease power and divide=−2x3+Csimplify\begin{aligned} \int-6x^2\,\mathrm{d}x&=\frac{-6}{3}x^3+C && \text{increase power and divide}\\ &=-2x^3+C && \text{simplify} \end{aligned}

Why does the sign stay negative? Differentiating −2x3-2x^3 produces −6x2-6x^2; there is no sign reversal in this inverse operation.

Answer. −2x3+C-2x^3+C.

Check. ddx(−2x3+C)=−6x2\dfrac{\mathrm{d}}{\mathrm{d}x}(-2x^3+C)=-6x^2.

A fractional coefficient is not a reason to use decimals. Leave it exact and let the new-power division simplify it.

Watch out

Wrong: ∫6x2 dx=3x3+C\int6x^2\,\mathrm{d}x=3x^3+C, found by dividing by the old power 22. Right: divide by the new power 33, giving 2x3+C2x^3+C. Differentiation exposes the wrong answer immediately: 3x33x^3 gives 9x29x^2.

Change the coefficient to a fraction for your turn.

Integrating a polynomial

Integrate a polynomial term by term, then write one constant of integration for the entire answer.

Continue this lesson.

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