Mathematics

Evaluate the following integral as a limit of sum:
$$\displaystyle \int_{0}^{1} \ (x+x^2) dx$$


SOLUTION
Let $$I=\displaystyle\int^{1}_0  (x+x^2) d x$$

As we know that

$$\displaystyle\int ^{b}_a f(x) d x=(b-a)\lim_{n\to \infty} \dfrac{1}{n}(f(a)+f(a+h)+f(a+2 h)+\cdots+f(a+(n-1)h))$$   where $$h=\dfrac{b-a}{n}$$ 

Here $$a=0,b=1,h=\dfrac{b-a}{n}=\dfrac{1-0}{n}=\dfrac{1}{n}$$ and $$f(x)=x^2+x$$

So $$I=(1-0)\displaystyle\lim_{n\to \infty} \dfrac{1}{n}(f(0)+f(h)+(f(2 h)+\cdots+f((n-1)h))$$

$$f(0)=0^2+0=0$$ 

$$f(h)=h^2+h=\dfrac{1}{n^2}+\dfrac{1}{n}$$

$$f((n-1)h)=(n-1)^2 h^2+(n-1)h=\dfrac{(n-1)^2}{n^2}+\dfrac{n-1}{n}$$

$$\therefore I=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(0+\dfrac{1}{n^2}+\dfrac{1}{n}+\dfrac{4}{n^2}+\dfrac{2}{n}+\cdots+\dfrac{(n-1)^2}{n^2}+\dfrac{n-1}{n}\bigg)$$

       $$=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(\bigg(\dfrac{1}{n^2}+\dfrac{2^2}{n^2}+\dfrac{3^2}{n^2}+\cdots+\dfrac{(n-1)^2}{n^2}\bigg)+\bigg(\dfrac{1}{n}+\dfrac{2}{n}+\dfrac{3}{n}+\cdots+\dfrac{n-1}{n}\bigg)\bigg)$$

       $$=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(\dfrac{1}{n^2}(1^2+2^2+3^2+\cdots+(n-1)^2+\dfrac{1}{n}(1+2+3+\cdots+(n-1))\bigg)$$

As we know that 

$$1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$$ and $$1^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2 n+1)}{6}$$

Hence $$I=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(\dfrac{1}{n^2}\times \dfrac{(n-1)n(2 n-1)}{6}+\dfrac{1}{n}\times \dfrac{(n-1)n}{2}\bigg)$$

             $$=\displaystyle\lim_{n\to \infty}\bigg(\dfrac{1}{n^2}\times \dfrac{(n-1)(2 n-1)}{6}+\dfrac{1}{n}\times \dfrac{(n-1)}{2}\bigg)$$

             $$=\displaystyle\lim_{n\to \infty}\bigg(\dfrac{1}{6} \bigg(1-\dfrac{1}{n}\bigg)\bigg(2-\dfrac{1}{n}\bigg)+\dfrac{1}{2}\bigg(1-\dfrac{1}{n}\bigg)\bigg)$$

            $$=\dfrac{1}{6}(1-0)(2-0)+\dfrac{1}{2}(1-0)$$

            $$=\dfrac{1}{3}+\dfrac{1}{2}$$

            $$=\dfrac{5}{6}$$
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