1.08a, 1.08b - Integration, antiderivatives and constants | OCR A-Level Mathematics A

Recover a function from its derivative, integrate powers and use a point to choose one curve from a family. Connect this reverse process to accumulated area and the fundamental theorem of calculus.

Recovering a family from a derivative

If you know every gradient of a curve, do you know where the curve is? Try the concrete case $y=x^2$ and $y=x^2+7$ . Both have gradient $2x$ everywhere. A complete map of the slopes still cannot tell you which curve you have. By the end, you will know exactly what extra information is needed.

You already know

Differentiating $x^n$ gives $nx^{n-1}$ . A constant differentiates to zero, and a vertical translation leaves every gradient unchanged.

That lost vertical position is the first thing integration must account for.

Antiderivative and indefinite integral

An antiderivative of $f(x)$ is a function $F(x)$ satisfying $F'(x)=f(x)$ . An indefinite integral represents the family of antiderivatives, written $\int f(x)\,\mathrm dx=F(x)+C$ . The integrand is $f(x)$ , and $\mathrm dx$ identifies the integration variable. For example, $x^2+7$ is an antiderivative of $2x$ ; $2x^2$ is not.

The constant of integration $C$ is any fixed real number on the interval considered. It is not another function of $x$ . Two antiderivatives on an interval differ by a constant: their difference has derivative zero. On disconnected intervals, each branch may have its own constant.

Vertical shifts have identical gradients

At $x=1$ , all three tangents have gradient $2$ , although their points of contact are at different heights. The picture contains infinitely many possible curves, even though only three are drawn.

Worked example 1: recover the family

Question. Find all functions with derivative $3x^2-3$ , on the real line.

Plan. Find one antiderivative, then restore the constant lost in differentiation.

\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^3-3x)&=3x^2-3 &&\text{check a candidate}\\ \frac{\mathrm d}{\mathrm dx}(x^3-3x+C)&=3x^2-3 &&\text{constant contributes zero} \end{aligned}

Why this step? Adding a constant changes the curve's height without changing any gradient.

Answer. The family is $y=x^3-3x+C$ , where $C\in\mathbb R$ .

Check. Both $C=7$ and $C=-2$ give the required derivative.

Now change only the constant term in the derivative. That changes a linear term in the antiderivative, not its arbitrary constant.

Worked example 2: a changed slope offset

Question. Find all functions with derivative $3x^2+2$ , on the real line.

Plan. Reverse the cubic and linear derivatives and keep an arbitrary constant.

\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^3+2x)&=3x^2+2 &&\text{reverse the derivatives}\\ y&=x^3+2x+C &&\text{include every vertical shift} \end{aligned}

Why this step? The term $2$ in the derivative comes from $2x$ , whereas $C$ contributes nothing to the derivative.

Answer. $\int(3x^2+2)\,\mathrm dx=x^3+2x+C$ .

Check. Differentiation gives $3x^2+2$ for every fixed $C$ .

A candidate must pass the derivative check before its constant is considered.

Watch out

Wrong: “The only antiderivative is $x^3+2x+5$ .” It is one member of the family. The complete indefinite integral is $x^3+2x+C$ .

Change the constant term in the derivative once more.

Reversing the power rule and combining terms

Increase the exponent by one, then divide by that new exponent. The division undoes the multiplier introduced by differentiation.

Power rule for integration

For constants $k,n$ , with $n\ne-1$ , on an interval where the real powers are defined and differentiable,

\int kx^n\,\mathrm dx=\frac{k}{n+1}x^{n+1}+C

This follows because differentiating the right side gives $kx^n$ . The sheet prints a more general reverse-chain power formula; this basic rule is a recall requirement.

Not in the formula sheet: learn it.

Sums, differences and constant multiples integrate term by term. Only one final constant is needed: several additive constants combine into one. A standalone constant also integrates: $\int7\,\mathrm dx=7x+C$ .

Worked example 3: roots and negative powers

Question. Find $\int(3x^2-4/\sqrt{x}+1/x^2)\,\mathrm dx$ , for $x>0$ .

Plan. Write the roots and reciprocals as powers, then integrate each term.

\begin{aligned} 3x^2-\frac4{\sqrt{x}}+\frac1{x^2}&=3x^2-4x^{-1/2}+x^{-2} &&\text{expose the exponents}\\ \int(3x^2-4x^{-1/2}+x^{-2})\,\mathrm dx &=x^3-8x^{1/2}-x^{-1}+C &&\text{divide by new powers} \end{aligned}

Why this step? The new exponent for $x^{-1/2}$ is $1/2$ , so dividing by it doubles the coefficient.

Answer. $x^3-8\sqrt{x}-1/x+C$ , for $x>0$ .

Check. Its derivative is $3x^2-4x^{-1/2}+x^{-2}$ , as required.

Change only the exponent of the final reciprocal term, from $-2$ to $-3$ . Its new exponent and coefficient both change.

Worked example 4: a steeper reciprocal

Question. Find $\int(3x^2-4/\sqrt{x}+1/x^3)\,\mathrm dx$ , for $x>0$ .

Plan. Keep the first two integrals and apply the power rule to $x^{-3}$ .

\begin{aligned} \int x^{-3}\,\mathrm dx&=\frac{x^{-2}}{-2} &&\text{new exponent is }-2\\ \int(3x^2-4x^{-1/2}+x^{-3})\,\mathrm dx &=x^3-8x^{1/2}-\frac12x^{-2}+C &&\text{combine terms} \end{aligned}

Why this step? Negative exponents increase towards zero when integrated; they do not become more negative.

Answer. $x^3-8\sqrt{x}-1/(2x^2)+C$ .

Check. Differentiating $-\tfrac12x^{-2}$ gives $x^{-3}$ .

Algebra may be needed first. Expanding $(x+2)^2$ gives an integral $x^3/3+2x^2+4x+C$ . The form $(x+2)^3/3+C$ is equivalent after absorbing its constant term into $C$ .

Watch out

The power rule does not apply to $x^{-1}$ : it would divide by zero. Nor can $x^0$ differentiate to $1/x$ . The reciprocal integral needs the logarithmic rule in a later lesson.

Use the same rewrite, with a different reciprocal exponent.

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1.08a, 1.08b - Integration, antiderivatives and constants

Recovering a family from a derivative

Reversing the power rule and combining terms

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