Area between curves and integration as the limit of a sumEdexcel A-Level Maths: Revision notes

If $f(x)\ge g(x)$ for $a\le x\le b$, the area between $y=f(x)$ and $y=g(x)$ is $\int_a^b\big(f(x)-g(x)\big)\mathrm{d}x.$ Find $a$ and $b$ by solving $f(x)=g(x)$. Example: $y=x^2-5x+6$ and $y=4-x^2$ meet where $2x^2-5x+2=0$, so $x=\frac12$ and $x=2$. At $x=1$ the second curve is higher, so area $=\int_{1/2}^2\left(-2x^2+5x-2\right)\mathrm{d}x=\frac98$.

Subtract first, then integrate: the difference $f-g$ already ignores the $x$-axis, so it works even if part of the region is below the axis. If the curves cross again inside the interval, split the region at every crossing, because the upper curve changes. Alternatively, area between a curve and a line $=$ area under the upper curve $-$ area under the lower one. This is quick when one boundary is a straight line, as the second area can be a triangle or trapezium.

For a curve with $x=x(t)$ and $y=y(t)$, replace $\mathrm{d}x$ by $\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t$: $A=\int y\,\mathrm{d}x=\int_{t_1}^{t_2}y(t)\,\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t.$ The limits must be values of $t$: convert the $x$-limits by solving $x(t)=a$ and $x(t)=b$. Example: $x=t^2$, $y=3t$, area from $x=0$ to $x=4$ ($t=0$ to $2$): $\int_0^26t^2\,\mathrm{d}t=16$. Example: $x=4\sin t$, $y=3\cos t$ gives $12\int_0^{\pi/2}\cos^2t\,\mathrm{d}t=3\pi$, using $\cos^2t=\frac12(1+\cos2t)$.

Divide the area under $y=f(x)$ from $a$ to $b$ into thin strips of width $\delta x$. Each strip is approximately a rectangle of area $f(x)\,\delta x$. The total area is approximately $\sum f(x)\,\delta x$, and this becomes exact as the strips get narrower: $\int_a^bf(x)\,\mathrm{d}x=\lim_{\delta x\to0}\sum_{x=a}^{x=b}f(x)\,\delta x.$ Example: for $y=x^2$ on $[0,3]$ with three left-hand rectangles, the sum is $0+1+4=5$, an underestimate of the exact area $\int_0^3x^2\,\mathrm{d}x=9$, because the curve is increasing.