Limit Definition of Integral

Sometimes, you’ll see an integral defined as a reimann sum, however, it’s expressed as an infinite limit as so →

∫abf(x)dx=lim⁡n→∞∑i=1nf(xi)Δx\int_{a}^{b}f\left(x\right)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f\left(x_{i}\right)\Delta x

Where →

  • Δx=b−an</span>, where n won’t be defined as a number\Delta x=\frac{b-a}{n}</span>\text{, where n won't be defined as a number}

  • xi=a+Δx⋅ix_{i}=a+\Delta x\cdot i

Here’s a few examples of how to convert from limit → integral and from integral → limit:

  1. Find the limit definition of this integral →∫28(x2+1)dx\int_2^8\left(x^2+1\right)dx

    1. First, we need to find our delta x → Δx=8−2n=6n\Delta x=\frac{8-2}{n}=\frac{6}{n}

    2. Next, we need to find our xi →xi=a+Δx⋅i=2+6n⋅ix_{i}=a+\Delta x\cdot i=2+\frac{6}{n}\cdot i

    3. Now, we take our function inside the integral and substitute the x’s for xi’s →f(xi)=(2+6ni)2+1f\left(x_{i}\right)=\left(2+\frac{6}{n}i\right)^2+1

    4. Now we need to put our limit and summation in front of the f(xi)dx →∫28(x2+1)dx=lim⁡n→∞∑i=1n[(2+6ni)2+1][6n]\int_2^8\left(x^2+1\right)dx=\lim_{n\to\infty}\sum_{i=1}^{n}\left\lbrack\left(2+\frac{6}{n}i\right)^2+1\right\rbrack\left\lbrack\frac{6}{n}\right\rbrack^{}

  2. Now, let’s turn this limit into an integral →lim⁡n→∞∑i=1n((1+2ni)3+ln⁡(1+2ni))(2n)\lim_{n\to\infty}\sum_{i=1}^{n}\left(\left(1+\frac{2}{n}i\right)^3+\ln\left(1+\frac{2}{n}i\right)\right)\left(\frac{2}{n}\right)

    1. First, we know that in our xi, the lower bound is always first, so a=1, now we use our delta x to find our b →Δx=2n,a=1…2=b−1…b=3\Delta x=\frac{2}{n},a=1\ldots2=b-1\ldots b=3

    2. Now, we substitute wherever there is an xi for an x →f(x)=x3+ln⁡xf\left(x\right)=x^3+\ln x

    3. Now we put our integral together →lim⁡n→∞∑i=1n((1+2ni)3+ln⁡(1+2ni))(2n)=∫13(x3+ln⁡x)dx\lim_{n\to\infty}\sum_{i=1}^{n}\left(\left(1+\frac{2}{n}i\right)^3+\ln\left(1+\frac{2}{n}i\right)\right)\left(\frac{2}{n}\right)=\int_1^3\left(x^3+\ln x\right)dx