Integration as the reverse of differentiationEdexcel A-Level Maths: Revision notes

Integration is the reverse of differentiation. If $F'(x)=f(x)$ then $\int f(x)\,dx=F(x)+c.$ This is the idea behind the Fundamental Theorem of Calculus: differentiation and integration undo each other, so differentiating the area function (or antiderivative) of $f$ returns $f$. Many functions have the same derivative, differing by a constant, so every indefinite integral needs a constant of integration $c$. Check any answer by differentiating it.

For any $n\neq-1$, $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+c.$ Add one to the power, then divide by the new power. The case $n=-1$ is excluded (it gives a logarithm, which comes later). Examples: $\int x^{3}dx=\frac{x^{4}}{4}+c$; $\int x^{-3}dx=-\frac12x^{-2}+c$; $\int\sqrt x\,dx=\frac23x^{3/2}+c$. Constant multiples and sums are integrated term by term: $\int\left(6x^{2}-4x+5\right)dx=2x^{3}-2x^{2}+5x+c$.

The rule only applies to powers of $x$, so rewrite first.

• Roots and reciprocals become powers: $\sqrt x=x^{1/2}$, $\frac1{x^{3}}=x^{-3}$.
• Expand brackets: $(x-1)^{2}=x^{2}-2x+1$.
• Divide each term separately. Example: $\frac{(x-1)^{2}}{\sqrt x}=x^{3/2}-2x^{1/2}+x^{-1/2}$, so $\int\frac{(x-1)^{2}}{\sqrt x}dx=\frac25x^{5/2}-\frac43x^{3/2}+2x^{1/2}+c$. Example: $\frac{(x+2)^{2}}{x^{4}}=x^{-2}+4x^{-3}+4x^{-4}$, which integrates to $-x^{-1}-2x^{-2}-\frac43x^{-3}+c$. Take care that a division does not produce an $x^{-1}$ term: $\frac{(x+2)^{2}}{x^{2}}$ contains $\frac4x$, which cannot be integrated with this rule.

Given $f'(x)$ and a point on the curve, integrate to get $f(x)+c$, then substitute the point to find $c$. Example: $\frac{dy}{dx}=6x^{2}-4x+5$ through $(1,7)$. $y=2x^{3}-2x^{2}+5x+c$, and $7=2-2+5+c$, so $c=2$ and $y=2x^{3}-2x^{2}+5x+2$. Example: $f'(x)=3x^{2}-\frac6{x^{3}}$ with $f(1)=2$ gives $f(x)=x^{3}+\frac3{x^{2}}-2$. At $x=2$ the point is $\left(2,\frac{27}4\right)$ and the gradient is $\frac{45}4$. Once you have the equation you can find stationary points: for $\frac{dy}{dx}=\frac{(x+2)^{2}}{x^{4}}$ the gradient is zero only at $x=-2$ but never negative, so there is no maximum or minimum there.