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Mathematics · Ch 9 — Applications of Integration

Limit Formula to Evaluate a Definite Integral

9.2.2

Limit Formula to Evaluate a Definite Integral

Rather than choose an arbitrary partition, divide [a,b][a,b] into nn equal subintervals [x0,x1],[x1,x2],…,[xn−1,xn][x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n], so that

x1−x0=x2−x1=⋯=xn−xn−1=b−an.x_1-x_0=x_2-x_1=\cdots=x_n-x_{n-1}=\dfrac{b-a}{n}.

Put h=b−anh=\dfrac{b-a}{n}; then xi=a+ihx_i=a+ih, i=1,2,…,ni=1,2,\ldots,n.

Deriving the limit formula. Using the right-end rule (ξi=xi\xi_i=x_i) from §9.2.1,

∫abf(x) dx=lim⁡n→∞∑i=1nf(xi)(xi−xi−1)=lim⁡n→∞b−an∑i=1nf ⁣(a+i⋅b−an),\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)(x_i-x_{i-1})=\lim_{n\to\infty}\dfrac{b-a}{n}\sum_{i=1}^n f\!\left(a+i\cdot\dfrac{b-a}{n}\right),

which (renaming the summation index i→ri\to r) gives the standard working formula

∫abf(x) dx=lim⁡n→∞b−an∑r=1nf ⁣(a+(b−a)rn).\boxed{\int_a^b f(x)\,dx=\lim_{n\to\infty}\dfrac{b-a}{n}\sum_{r=1}^n f\!\left(a+\dfrac{(b-a)r}{n}\right).}

A short computation (isolating the r=0r=0 term, which is b−anf(a)→0\frac{b-a}{n}f(a)\to0 as n→∞n\to\infty) shows the sum may equivalently start at r=0r=0: lim⁡n→∞b−an∑r=0nf ⁣(a+(b−a)rn)\displaystyle\lim_{n\to\infty}\dfrac{b-a}{n}\sum_{r=0}^n f\!\left(a+\dfrac{(b-a)r}{n}\right) gives the same value.

Special case a=0, b=1a=0,\ b=1. The formula collapses to

∫01f(x) dx=lim⁡n→∞1n∑r=1nf ⁣(rn).\int_0^1 f(x)\,dx=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n f\!\left(\dfrac{r}{n}\right). …

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