Area between curvesAQA A-Level Maths: Revision notes

The definite integral $\int_a^b f(x)\,dx$ gives the area between the curve $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$, provided the curve is above the axis. Where the curve is below the $x$-axis the integral is negative, so the area is its modulus. Evaluate with the limits: $\int_a^b f(x)\,dx=F(b)-F(a)$, where $F$ is an antiderivative of $f$.

If $f(x)\geq g(x)$ for all $x$ in $[a,b]$, the area between the graphs is $A=\int_a^b\big[f(x)-g(x)\big]\,dx,$ that is, upper minus lower. This is the area under $f$ minus the area under $g$, so it also works when part of the region is below the $x$-axis, as long as $f\geq g$ throughout. A straight line is just another curve, so the same rule applies to a curve and a line.

The limits are usually the $x$-coordinates where the two graphs meet. Set $f(x)=g(x)$, rearrange to $0$ and solve. To decide which graph is above, substitute a value of $x$ between the intersections (for example the midpoint) into both; the larger $y$-value belongs to the upper graph. A quick sketch helps, but the substitution is the proof.

Find the area enclosed by $y=x^2+1$ and $y=x+3$. Intersections: $x^2+1=x+3\Rightarrow x^2-x-2=0\Rightarrow(x+1)(x-2)=0$, so $x=-1,\,2$. At $x=0$ the line gives $3$ and the curve gives $1$, so the line is above. $A=\int_{-1}^{2}(2+x-x^2)\,dx=\left[2x+\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^{2}=\frac{10}{3}-\left(-\frac76\right)=\frac92.$ Take care substituting a negative limit: $2(-1)+\frac{(-1)^2}{2}-\frac{(-1)^3}{3}=-2+\frac12+\frac13=-\frac76$.

If the graphs cross between the limits, the upper graph changes, so one integral of $f-g$ would let positive and negative areas cancel. Split the region at each crossing and use upper minus lower on each piece. Example: $y=x^3$ and $y=x$ meet where $x^3-x=0$, at $x=-1,0,1$. For $0<x<1$ the line is above, and for $-1<x<0$ the curve is above. By symmetry, $A=2\int_0^1(x-x^3)\,dx=2\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1=\frac12.$ A single integral from $-1$ to $1$ gives $0$, which is wrong.

Some questions combine ideas: find the area under a curve first, then use it with the area between a curve and a line. For example, if $C$ and the $x$-axis bound an area of $36$ and the region between $C$ and a line cuts off $\frac{125}{6}$ of it, the proportion is $\frac{125/6}{36}=\frac{125}{216}$. Keep exact fractions until the end, and give decimals only if the question asks.