Question

Evaluate the following as the limit of sum :

${\int }_0^2(x+4)dx$

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is D $10$

We have,

${\int }_a^{b}f\left(x\right)dx=li{m}_{h\to 0}h\left[f\right(a)+f(a+h)+f(a+2h)+....+f(a+(n-a)h\left)\right]$

Here, $a=0,b=2,f\left(x\right)=x+4$ and $h=\frac{2-0}{n}=\frac{2}{n}$

Therefore,

$I={\int }_0^2(x+4)dx=li{m}_{h\to 0}h\left[f\right(0)+f(0+h)+f(0+2h)+...+f(0+(n-1)h\left)\right]$

$I=li{m}_{h\to 0}h\left[\right(0+4)+(h+4)+(2h+4)+...+(n-1)h+4)]$

$I=li{m}_{h\to 0}h[4n+h(1+2+3+....+(n-1)]$

$I=li{m}_{h\to 0}h[4n+h\frac{n(n-1)}{2}]$

$I=li{m}_{n\to \infty }\frac{2}{n}[4n+\frac{2}{n}\times \frac{n(n-1)}{2}]$

$I=li{m}_{n\to \infty }[8+2(1-\frac{1}{n}\left)\right]=8+2(1-0)=10$