Pure 8.1 - Fundamental Theorem of Calculus | Edexcel A-Level Mathematics

Can a gradient function tell you exactly where a curve is? The Fundamental Theorem of Calculus links differentiation to accumulation, and integration recovers a function from its derivative. You will see why a constant is needed, use a point to find one particular curve, and evaluate definite integrals by subtracting endpoint values.

The inverse link

If every point on a curve has gradient $x$ , can you recover the curve? Start with $y=\frac{x^2}{2}$ . Differentiation gives $x$ , but so do $\frac{x^2}{2}+5$ and $\frac{x^2}{2}-3$ . You can recover the shape, yet its height remains unknown.

You already know

$\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=nx^{n-1}$ , and a constant differentiates to zero. Differentiation acts term by term on a sum.

Running these derivative facts backwards is the starting point for integration. For example, $\frac{\mathrm{d}}{\mathrm{d}x}(2x^2)=4x$ , so $2x^2$ is one answer to “what differentiates to $4x$ ?”

Antiderivative

An antiderivative of $f$ on an interval is a function $F$ such that $F'(x)=f(x)$ throughout that interval. For example, $2x^2+1$ and $2x^2-6$ are antiderivatives of $4x$ . Not an example: $4x^2$ differentiates to $8x$ .

The whole family is written using an arbitrary real constant $C$ . On a connected interval, any two antiderivatives of the same function differ by a constant: their difference has derivative zero.

Indefinite integration

If $F'(x)=f(x)$ on an interval, then

\int f(x)\,\mathrm{d}x=F(x)+C

where $C$ is an arbitrary constant on that interval.

Not in the formula booklet: learn it.

The integral sign names the reverse process; an indefinite integral describes a family rather than a single number. The picture makes the missing information visible.

[DIAGRAM: asset_slug: p1_8_1_know_and_use_the_fundamental_theorem_of_calculus__diagram_01; asset_name: Antiderivatives share gradients; caption: Vertical shifts change the height but preserve the gradient at each x-value; description: Two panels connect F(x)=x squared over two plus C with derivative x, and show three vertically shifted parabolas with parallel tangent segments at x=1]

A surprising consequence is that a complete list of gradients still cannot distinguish curves separated vertically by a million units. One height measurement can do what all those gradients cannot.

Key idea

Differentiation loses an additive constant; indefinite integration records that missing information as $+C$ .

Try distinguishing a family of answers from a single incorrect reversal.

Why plus C matters

The three functions $3x^2+7x-4$ , $3x^2+7x$ and $3x^2+7x+11$ all differentiate to $6x+7$ . The complete answer to integrating $6x+7$ must therefore be $3x^2+7x+C$ .

For now, choose terms by reversing familiar derivative facts. The next lesson develops the general power rule.

Worked example 1: reversing two derivative facts

Question. Find $\int(10x^4-3)\,\mathrm{d}x$ .

Plan. Find a term whose derivative is $10x^4$ , do the same for $-3$ , then add the missing constant.

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}(2x^5)&=10x^4 && \text{reverse the power derivative}\\ \frac{\mathrm{d}}{\mathrm{d}x}(-3x)&=-3 && \text{reverse a constant gradient}\\ \int(10x^4-3)\,\mathrm{d}x&=2x^5-3x+C && \text{combine the terms} \end{aligned}

Why this step? $-3$ is a constant gradient, so its antiderivative is $-3x$ , not another constant.

Answer. $2x^5-3x+C$ .

Check. Differentiating gives $10x^4-3+0$ , exactly the integrand.

Change just the coefficient of the power term, from $10$ to $15$ . The constant-gradient term and the need for $+C$ stay the same.

Worked example 2: changing one coefficient

Question. Find $\int(15x^4-3)\,\mathrm{d}x$ .

Plan. Reverse each derivative, allowing for the changed coefficient.

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}(3x^5)&=15x^4 && \text{coefficient }3\times5=15\\ \int(15x^4-3)\,\mathrm{d}x&=3x^5-3x+C && \text{include the whole family} \end{aligned}

Why this step? The coefficient must be $3$ , because differentiating the fifth power multiplies it by $5$ .

Answer. $3x^5-3x+C$ .

Check. $\frac{\mathrm{d}}{\mathrm{d}x}(3x^5-3x+C)=15x^4-3$ .

Writing a complete family matters even when you have found a correct individual antiderivative.

Watch out

Wrong: $\int(6x+7)\,\mathrm{d}x=3x^2+7x$ as a complete indefinite integral. Right: $3x^2+7x+C$ . The expression without $C$ is only one member of the family.

Try another coefficient; include the constant in the final expression.

Notation and variable

In $\int6t^2\,\mathrm{d}t$ , the $\mathrm{d}t$ tells you to integrate with respect to $t$ . Since differentiating $2t^3$ gives $6t^2$ , the answer is $2t^3+C$ . Renaming the variable does not change the reverse process.

Integrand and integration variable

In $\int f(x)\,\mathrm{d}x$ , $f(x)$ is the integrand and $x$ is the integration variable. The differential $\mathrm{d}x$ identifies that variable. An integral without endpoint limits is indefinite.

Term-by-term reversal works in any named variable. For example,

\begin{aligned} \int(3x^2-4x+5)\,\mathrm{d}x&=x^3-2x^2+5x+C\\ \int(3t^2-4t+5)\,\mathrm{d}t&=t^3-2t^2+5t+C \end{aligned}

Differentiating each line with respect to its own variable checks every coefficient. These term-by-term rules are not in the formula booklet: learn them.

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Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 8.1 - Fundamental Theorem of Calculus

The inverse link

Why plus C matters

Notation and variable

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