Integration as the limit of a sumAQA A-Level Maths: Revision notes

To find the area under a curve $y=f(x)$ from $x=a$ to $x=b$, divide the region into thin vertical strips of equal width $\delta x$ and draw a rectangle on each. A rectangle at position $x$ has height $f(x)$, so its area is $f(x)\,\delta x$, and the total area is approximately $\sum f(x)\,\delta x$. Choosing the height from the left-hand end or the right-hand end gives two different sums. For $y=x^2$ on $[0,3]$ with three strips of width 1, left-hand ends give $0+1+4=5$ and right-hand ends give $1+4+9=14$, while the exact area is 9.

If $f$ is increasing the left-hand rectangles lie below the curve (an underestimate) and the right-hand rectangles reach above it (an overestimate). If $f$ is decreasing it is the other way round. The exact area lies between the two sums, and the error is the difference between the sum and the exact area, e.g. $\frac{9-5}{9}=44.4\%$ for the three left-hand rectangles on $y=x^2$. Using more, narrower strips reduces the error.

As the strip width $\delta x\to0$ the rectangles fit the curve more and more closely: $\int_a^bf(x)\,dx=\lim_{\delta x\to0}\sum_{x=a}^{b}f(x)\,\delta x.$ The integral sign $\int$ is a stretched 'S' for sum, and $dx$ is what $\delta x$ becomes. This is why a definite integral gives area when $f(x)\ge0$. To go from a sum to an integral, read off the limits from the first and last values of $x$ in the sum and the function $f(x)$ multiplying $\delta x$: $\lim\sum_{x=1}^{4}(2x+1)\delta x=\int_1^4(2x+1)dx=18$.

With $n$ strips of equal width on $[a,b]$ the width is $\frac{b-a}{n}$ and the $r$th right-hand end is $x=a+r\frac{b-a}{n}$. For $y=3x$ on $[0,2]$ the width is $\frac2n$, the heights are $\frac{6r}{n}$, and using $\sum_{r=1}^nr=\frac{n(n+1)}{2}$: $\sum_{r=1}^n\frac2n\cdot\frac{6r}{n}=\frac{12}{n^2}\cdot\frac{n(n+1)}{2}=6+\frac6n.$ As $n\to\infty$, $\frac6n\to0$, so the sum tends to $6$, which equals $\int_0^23x\,dx$. The term $\frac6n$ is the overestimate and shows how quickly the error falls.

Any quantity made of many small contributions can be written as a limit of a sum. For a rod of density $\rho(x)$ kg per metre, a piece of length $\delta x$ has mass about $\rho(x)\,\delta x$, so the total mass is $\lim\sum\rho(x)\delta x=\int\rho(x)dx$. With $\rho=2+\frac x2$ on $0\le x\le4$: $\int_0^4\left(2+\frac x2\right)dx=\left[2x+\frac{x^2}{4}\right]_0^4=12$ kg. The same idea gives the mass of the first $k$ metres, $2k+\frac{k^2}{4}$, which can be set equal to a given value and solved, rejecting any root outside the domain.