Question

integration 0 to 2 (2x^2+5x+1) as a limit of sum​

Answer 1

Answer:

The answer is completed

Answer 2

The value of the given integration is  $\int\limits_0^2 (2x^2+5x+1)dx=\frac{52}{3}$

Step-by-step explanation:

Given that integration 0 to 2 (2x^2+5x+1) as a limit of sum​

It can be written as

$\int\limits_0^2 (2x^2+5x+1)dx$

To find the given integration :

• $\int\limits_0^2 (2x^2+5x+1)dx$
• $=(\frac{2x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}+x+c)^2_0$
• $=(\frac{2x^3}{3}+\frac{5x^2}{2}+x+c)^2_0$ ( by using the formula $\int x^ndx=\frac{x^{n+1}}{n+1}$  and $\int dx=x+c$ )
• $=\frac{2(2)^3}{3}+\frac{5(2)^2}{2}+2-(\frac{2(0)^3}{3}+\frac{5(0)^2}{2}+0)$

• $=\frac{16}{3}+\frac{20}{2}+2-0$
• $=\frac{16}{3}+10+2$
• $=\frac{16}{3}+12$
• $=\frac{16+12(3)}{3}$

• $=\frac{16+36}{3}$
• $=\frac{52}{3}$

Therefore $\int\limits_0^2 (2x^2+5x+1)dx=\frac{52}{3}$

The value of the given integration is  $\int\limits_0^2 (2x^2+5x+1)dx=\frac{52}{3}$