Definite integrals and area under a curveEdexcel A-Level Maths: Revision notes

A definite integral has limits $a$ and $b$. Integrate, then substitute the limits and subtract: $\int_a^bf(x)\,\mathrm{d}x=\big[F(x)\big]_a^b=F(b)-F(a).$ No constant $c$ is needed because it cancels. Two useful properties: swapping the limits changes the sign, $\int_b^af=-\int_a^bf$, and an interval can be split, $\int_a^cf=\int_a^bf+\int_b^cf$. Example: $\int_0^4\left(6x-x^2\right)\mathrm{d}x=\left[3x^2-\frac{x^3}{3}\right]_0^4=48-\frac{64}{3}=\frac{80}{3}$.

If $y=f(x)\ge0$ for $a\le x\le b$, the area between the curve, the $x$-axis and the lines $x=a$ and $x=b$ is $\int_a^by\,\mathrm{d}x$. Find where the curve meets the axis by solving $y=0$; these roots often give the limits. Sketch the curve first so that you can see which parts are above and below the axis.

Where the curve is below the $x$-axis, $y<0$ so the integral is negative. The integral is then the negative of the area. For $y=x^2-4x+3$, $\int_1^3y\,\mathrm{d}x=-\frac43$ so the area is $\frac43$. If a region lies partly above and partly below the axis, integrating across the root lets the positive and negative parts cancel. Instead split at each root, find each integral, and add the positive areas. For $y=x^2-4x+3$ between $x=0$ and $x=4$ the integrals are $\frac43,-\frac43,\frac43$, so the total area is $4$ although $\int_0^4y\,\mathrm{d}x=\frac43$.

For a region between a curve and a line, the area is $\int_a^b\left(y_{\text{top}}-y_{\text{bottom}}\right)\mathrm{d}x$, where $a$ and $b$ are the $x$-coordinates of the intersection points. Example: $y=6x-x^2$ and $y=2x$. Solving $6x-x^2=2x$ gives $x=0$ and $x=4$. Between them the curve is above the line, so area $=\int_0^4\left(4x-x^2\right)\mathrm{d}x=\left[2x^2-\frac{x^3}{3}\right]_0^4=\frac{32}{3}$. An equivalent method is the area under the curve minus the area under the line: $\frac{80}{3}-16=\frac{32}{3}$.