Area between a curve and a line or two curvesEdexcel International A Level Maths: Revision notes

For a region bounded above by $y=f(x)$ and below by $y=g(x)$ between $x=a$ and $x=b$: $\text{Area}=\int_a^b\left[f(x)-g(x)\right]dx,$ upper minus lower. This works even when part of the region is below the $x$-axis, because the difference of the heights is still positive. Example: $y=6x-x^2$ and $y=2x$ meet where $x^2-4x=0$, at $x=0$ and $x=4$. The curve is above the line, so area $=\int_0^4(4x-x^2)\,dx=\left[2x^2-\frac{x^3}{3}\right]_0^4=\frac{32}{3}$.

The limits come from the points where the graphs meet. Equate the two expressions, rearrange to $0$ and solve (factorise or use the quadratic formula). Check each $x$-value gives a point on both graphs. Example: $y=x^2$ and $y=x+2$ give $x^2-x-2=0$, so $x=-1$ and $x=2$. To find which graph is higher, test a value between the limits, e.g. $x=0$: line $y=2$, curve $y=0$, so the line is above.

The same rule applies: area $=\int_a^b\left[\text{upper}-\text{lower}\right]dx$, with $a$ and $b$ the $x$-coordinates of the intersections that bound the region. Example: $y=x^2$ and $y=8-x^2$ meet at $x=\pm2$, and $8-x^2$ is above, so area $=\int_{-2}^{2}(8-2x^2)\,dx=\left[8x-\frac{2x^3}{3}\right]_{-2}^{2}=\frac{64}{3}$. Where the region is symmetrical you may double the integral from $0$, but only if you can justify the symmetry.

The area between a line and a curve is also (area under the upper graph) $-$ (area under the lower graph), each found with $\int y\,dx$ between the same limits. If the line forms a trapezium or triangle, use geometry for its area. Example: the area under $y=x+2$ from $x=-1$ to $x=2$ is a trapezium with parallel sides $1$ and $4$ and width $3$: $\frac{15}{2}$. The area under $y=x^2$ is $3$. The region between them is $\frac{15}{2}-3=\frac92$.

Find the area bounded by $y=x^2$ and $y=2x+3$. Intersections: $x^2=2x+3$, so $x^2-2x-3=(x-3)(x+1)=0$, giving $x=-1$ and $x=3$. At $x=0$ the line ($3$) is above the curve ($0$). Area $=\int_{-1}^{3}(2x+3-x^2)\,dx=\left[x^2+3x-\frac{x^3}{3}\right]_{-1}^{3}=9-\left(-\frac53\right)=\frac{32}{3}$.