Question

Express ${\int }_0^4{x}^{3}dx$ as limit of sum and thus evaluate it.

Answer 1

${\int }_0^4{x}^{3}dx$

Evaluating integral as limit of sum

${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n⟶\infty }{lim}\frac{1}{n}[f\left(a\right)+f(a+h)+...f(a+(n-1)h)]h=\frac{b-a}{n}$

Here, $h=\frac{4-0}{n}=\frac{4}{n}=4\underset{n⟶\infty }{lim}\frac{1}{n}[f\left(0\right)+f\left(\frac{4}{n}\right)+....f\left(\frac{(n-1)4}{n}\right)]=4\underset{n⟶\infty }{lim}\frac{1}{n}[0+{\left(\frac{4}{n}\right)}^3+{\left(\frac{2.4}{n}\right)}^3+....{\left(\frac{(n-1)4}{n}\right)}^3]=4\underset{n⟶\infty }{lim}\frac{1}{n}\left[{\left(\frac{4}{n}\right)}^3(1^3+2^3+...{(n-1)}^3)\right]$

We know, $=1^3+2^3+3^3+....nterms={\left(\frac{n(n+1)}{2}\right)}^2=4\underset{n⟶\infty }{lim}\frac{1}{n}{\left(\frac{4}{n}\right)}^3{\left(\frac{n(n+1)}{2}\right)}^2=4^4\underset{n⟶\infty }{lim}\frac{1}{n}\frac{1}{n^4}\times \frac{n^2{(n-1)}^2}{4}=4^3\underset{n⟶\infty }{lim}\frac{1}{n}\frac{n^2-2n+1}{n^2}=4^3\underset{n⟶\infty }{lim}(1-\frac{2}{n}+\frac{1}{n^2})=4^3=64$