H1, H2 - Antiderivatives and the fundamental theorem | AQA A-Level Mathematics

Recover a function from its derivative, integrate rational powers and use an antiderivative to measure accumulated change. H2 supplies the integration methods; H1 explains why they also evaluate a definite integral.

Antiderivatives and constants

If you know a curve's gradient everywhere, can you recover the curve itself? The curves $y=x^2$ , $y=x^2+2$ and $y=x^2-2$ all have derivative $2x$ . Their slopes agree, but their heights do not.

You already know

Differentiating a constant gives zero. Differentiating $x^n$ gives $nx^{n-1}$ , wherever the expression is real and differentiable. These differentiation rules are not in the formula booklet: learn them.

Read differentiation backwards: $x^2$ is one function whose derivative is $2x$ .

Antiderivative

An antiderivative of $f(x)$ on an interval is a function $F(x)$ satisfying $F'(x)=f(x)$ throughout that interval. For example, $x^2+7$ is an antiderivative of $2x$ ; $x^2+7x$ is not, because its derivative is $2x+7$ .

Adding any constant leaves the derivative unchanged. These three members of an infinite family all have gradient $2$ at $x=1$ .

Three antiderivatives of 2x, shifted vertically, with parallel tangents of gradient 2 at x equals 1

The indefinite integral records the whole family. The $dx$ identifies the variable of integration.

The constant of integration

If $F'=f$ on an interval, then

\int f(x)\,dx=F(x)+C

where $C$ is an arbitrary constant. Any two antiderivatives on that interval differ by a constant: their difference has derivative zero.

Not in the formula booklet: learn it.

A point on the curve selects one member. Integrate first, then use the point to find $C$ .

Worked example 1: selecting the curve

Question. A function satisfies $f'(x)=4x-3$ and $f(2)=7$ . Find $f(x)$ .

Plan. Reverse differentiation term by term, then substitute the given value.

\begin{aligned} f(x)&=2x^2-3x+C && \text{integrate both terms} \\ 7&=2(2)^2-3(2)+C && \text{use }f(2)=7 \\ C&=5 && \text{solve }7=2+C \end{aligned}

Why this step? The derivative determines the variable terms; the point fixes their vertical shift.

Answer. $f(x)=2x^2-3x+5$ .

Check. $f'(x)=4x-3$ and $f(2)=8-6+5=7$ .

Change only the given height from $7$ to $11$ . The derivative stays the same, so only the constant should change.

Worked example 2: a different height

Question. A function satisfies $g'(x)=4x-3$ and $g(2)=11$ . Find $g(x)$ .

Plan. Use the same antiderivative family and determine its new constant.

\begin{aligned} g(x)&=2x^2-3x+C && \text{same derivative} \\ 11&=8-6+C && \text{substitute }x=2 \\ C&=9 && \text{subtract }2 \end{aligned}

Why this step? Raising the specified height by $4$ raises the whole curve by $4$ , without changing its gradient.

Answer. $g(x)=2x^2-3x+9$ .

Check. $g'(x)=4x-3$ , $g(2)=11$ and $g(x)-f(x)=4$ .

The arbitrary constant is added to the function, rather than multiplied by the variable.

Watch out

Wrong: $\int 2x\,dx=x^2+Cx$ . Its derivative is $2x+C$ , which changes the integrand. Right: $x^2+C$ , whose derivative is $2x$ for every constant $C$ .

Change the specified height once more. Your variable terms should still be identical.

Integrating rational powers

Writing a root or denominator as a power makes its antiderivative easier to recognise: $1/\sqrt{x}=x^{-1/2}$ , while $1/x^3=x^{-3}$ .

For instance, differentiating $2\sqrt{x}=2x^{1/2}$ gives $x^{-1/2}$ . The new exponent is one greater than the old one, and its coefficient cancels the factor introduced by differentiation.

The power rule for integration

For rational $n\ne-1$ , on an interval where the functions are real and differentiable,

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

Integrate sums and differences term by term, retaining constant multipliers. In particular, $\int k\,dx=kx+C$ .

Not in the formula booklet: learn it.

The reason is $\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n$ . When $n=-1$ , the denominator would be zero: the reciprocal needs the next lesson's logarithmic rule. For fractional or negative powers, $x>0$ is a safe common domain. Non-negative integer powers work for every real $x$ ; other domains must be checked. Separate intervals of a disconnected domain may have independent constants; a point on one interval does not fix the constant on another.

Worked example 3: roots and negative powers

Question. For $x>0$ , find $\displaystyle\int\left(3x^2-\frac4{\sqrt{x}}+\frac5{x^3}\right)\,dx$ .

Plan. Write the quotients as powers, then increase each exponent and divide by its new value.

\begin{aligned} I&=\int(3x^2-4x^{-1/2}+5x^{-3})\,dx && \text{write as powers} \\ &=\frac{3x^3}{3}-\frac{4x^{1/2}}{1/2}+\frac{5x^{-2}}{-2}+C && \text{apply the rule} \\ &=x^3-8\sqrt{x}-\frac5{2x^2}+C && \text{simplify coefficients} \end{aligned}

Why this step? The last new exponent is negative, so dividing by it changes that term's sign.

Answer. $x^3-8\sqrt{x}-\frac5{2x^2}+C$ , for $x>0$ .

Check. Its derivative is $3x^2-4x^{-1/2}+5x^{-3}$ , the original integrand.

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H1, H2 - Antiderivatives and the fundamental theorem

Antiderivatives and constants

Integrating rational powers

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