20.997
1) Let's start by dividing from 0 to 4 into n equal subintervals:
![n=4,x=((4-0))/(4)=1](https://img.qammunity.org/2023/formulas/mathematics/college/1mqal267tuwin223u3gaq6wnzu5417nzrd.png)
So the width of each rectangle is going to be 1
2) The next step is to calculate the area below the curve using the right endpoints using the following intervals:
![\lbrack0,1\rbrack,\lbrack1,2\rbrack,\lbrack2,3\rbrack,\lbrack3,4\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/cn5qup4wghqn0b5mah6j60xkj6rw64d2iu.png)
And plug each endpoint into that we have:
![\begin{gathered} R_4=f(x_1)\cdot x+f(x_2)x+f(x_3)x+f(x_4)x \\ R_4=f(1)\cdot1+f(2)\cdot1+f(3)\cdot1+f(4)\cdot1 \\ R_4=(\sqrt[]{9(1)+6})\cdot1+(\sqrt[]{9(2)+6})\cdot1+(\sqrt[]{9(3)+6})\cdot1+(\sqrt[]{9(4)+6})\cdot1 \\ R_4=\sqrt[]{15}+2\sqrt[]{6}+\sqrt[]{33}+\sqrt[]{42}\approx20.997 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p8e58clmsmp1un54xvxbmdywcfyisui472.png)
Note that since each rectangle has a width of 1 unit, we could plug it into that.
3) Hence, the answer is approximately 20.997