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

Integration is the reverse of differentiation. If $F'(x)=f(x)$ then $\int f(x)\,dx=F(x)+c$, where $c$ is the constant of integration. This is the Fundamental Theorem of Calculus in its indefinite form: differentiating an integral returns the original function, $\frac{d}{dx}\int f(x)\,dx=f(x)$, and integrating a derivative returns the function, $\int f'(x)\,dx=f(x)+c$. Because the derivative of a constant is zero, many functions share the same derivative, which is why $+c$ is needed.

For $n\neq-1$: $\int x^n\,dx=\frac{x^{n+1}}{n+1}+c.$ Add one to the power, then divide by the new power. The rule holds for negative and fractional $n$: $\int x^{-2}dx=-x^{-1}+c=-\frac1x+c$ and $\int x^{\frac12}dx=\frac{x^{3/2}}{3/2}+c=\frac23x^{\frac32}+c$. The rule fails for $n=-1$, because the denominator $n+1$ would be zero. Check any answer by differentiating it.

Integrate term by term, and constants multiply through: $\int\left(af(x)\pm bg(x)\right)dx=a\int f(x)\,dx\pm b\int g(x)\,dx$. For $\frac{dy}{dx}=6x^2-4x+5$, $y=2x^3-2x^2+5x+c$. A constant $k$ integrates to $kx$. Only one constant $c$ is needed for the whole expression.

There is no rule for integrating a product or quotient directly, so rewrite the expression as a sum of powers first: expand brackets, write roots as powers ($\sqrt x=x^{\frac12}$) and split fractions. For $\frac{(x^2+3)^2}{x^2}$, expand to $\frac{x^4+6x^2+9}{x^2}=x^2+6+9x^{-2}$, then integrate: $\frac{x^3}{3}+6x-\frac9x+c$. Similarly $\frac8{x^2}=8x^{-2}$ and $3\sqrt x=3x^{\frac12}$.

A single point on the curve fixes the constant. Substitute its coordinates into the integrated equation and solve for $c$. For $\frac{dy}{dx}=6x^2-4x+5$ through $(1,3)$: $y=2x^3-2x^2+5x+c$ gives $2-2+5+c=3$, so $c=-2$. Curves with the same gradient function differ only by the constant, so they are vertical translations of each other. With an unknown constant in the gradient, such as $\frac{dy}{dx}=2kx-3$, use one point to find $c$ and another to find $k$.