Question

Find ${\int }_0^2(x^2+1)dx$ as the limit of a sum.

• A. $\frac{4}{3}$
• B. $\frac{14}{3}$
• C. $\frac{14}{5}$
• D. None of these

Answer 1

The correct option is B $\frac{14}{3}$

We know that

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

Putting $a=0,b=2,h=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$

in ${\int }_0^2{x}^2+1dx$

$I=(2-0)\underset{n\to \infty }{lim}\frac{1}{n}\left(f\right(0)+f(n)+f(2n)+...+f((n-1)h\left)\right)$

$f\left(0\right)=1$

$f\left(h\right)=h^2+1$

$=\left(\frac{4}{n^2}\right)+1$

$f\left(\right(n-1\left)h\right)=(n-1{)}^2\times \frac{4}{n^2}+1$

$\therefore I=2\underset{n\to \infty }{lim}\frac{1}{n}\left(\right(1+1+...ntimes)+(0+\frac{4}{n^2}+\frac{16}{n^2}+...+\frac{(n-1{)}^2}{n^2}))$

$=2\underset{n\to \infty }{lim}\frac{1}{n}(n+\frac{4}{n}\frac{(n-1)n(2n-1)}{6})$

$=2\underset{n\to \infty }{lim}(1+\frac{2}{3}(1-\frac{1}{n})(2-\frac{1}{n}))$

$=2\times (1+\frac{4}{3})=\frac{14}{3}$