Question

$\int^2 _1(4x^2-1)dx$
find the following integrals as the limits of sums ​

Answer 1

Given :

• $\sf {\int\limits^2_1 (4x^2-1) dx}$

Formula Applied :

$\sf {\int\limits^b_a {f(x)} \, dx = \lim_{n \to \infty} \dfrac{b-a}{n} \sum_{r=1}^{n}f \big(a+(b-a)\dfrac{r}{n}\big)}$

• $\sf f(x)=4x^2-1$
• $\sf a = 1$
• $\sf b=2$

Solution :

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{2n-1}{n} \sum_{r=1}^{n} \bigg(1+\dfrac{r}{n}\bigg)}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} \bigg[4\bigg(1+\dfrac{r}{n}\bigg)^2-1\bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} \bigg[4\bigg(1+\dfrac{2r}{n}+\dfrac{r^2}{n^2}\bigg)-1\bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} \bigg[4+\dfrac{8r}{n}+\dfrac{4r^2}{n^2}-1\bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} \bigg[\dfrac{8r}{n}+\dfrac{4r^2}{n^2}+3\bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \bigg[\sum_{r=1}^n \dfrac{4r^2}{n^2}+\sum_{r=1}^{n}\dfrac{8r}{n}+\sum_{r=1}^{n} 3 \bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \bigg[4 \times \dfrac{n(n+1)(2n+1)}{6n^2}+\dfrac{8}{n} \times \dfrac{n(n+1)}{2}+3n\bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \dfrac{1}{n} \bigg[4 \times \dfrac{n^3\Big(1+\dfrac{1}{n}\Big)\Big(2 \times \dfrac{1}{n}\Big)}{6n^2}+\dfrac{8n}{2}\bigg(1+\dfrac{1}{n}\bigg)+3n\bigg]}$

${\Rightarrow \sf \lim_{n \to \infty} \bigg[\dfrac{2}{3}\times \bigg(1+\dfrac{1}{n}\bigg)\bigg(2+\dfrac{1}{n}\bigg)+ 4 \bigg(1+\dfrac{1}{n}\bigg)+3\bigg]}$

${\Rightarrow \sf \Bigg[\dfrac{2}{3}\times \bigg(1+\dfrac{1}{\Big(\dfrac{1}{0}\Big)}\bigg)\bigg(2+\dfrac{1}{\Big(\dfrac{1}{0}\Big)}\bigg)+ 4 \bigg(1+\dfrac{1}{\Big(\dfrac{1}{0}\Big)}\bigg)+3\Bigg]}$

${\Rightarrow \sf \bigg[\dfrac{2}{3}\Big(1\Big)\Big(2\Big)+4\Big(1\Big)+3\bigg] \Rightarrow \dfrac{4}{3}+4+3}$

${\Rightarrow \sf \dfrac{4}{3}+7 \Rightarrow \dfrac{4+21}{3}\Rightarrow \dfrac{25}{3}}$

Required Answer :

$\boxed{\boxed{\sf \int\limits^2_1 (4x^2-1) dx = \dfrac{25}{3}}}$