The interval has size 1 so Δ x = 1 n {\displaystyle \Delta x={\frac {1}{n}}} . The interval is [ 0 , 1 ] {\displaystyle [0,1]} so x i = i n {\displaystyle x_{i}={\frac {i}{n}}} . In this case f ( x ) = 6 − 3 x 2 {\displaystyle f(x)=6-3x^{2}} so the right Riemann sum is:
R n = ∑ i = 1 n ( 6 − 3 ( i n ) 2 ) ( 1 n ) = 6 n ∑ i = 1 n 1 − 3 n 3 ∑ i = 1 n i 2 = 6 n ( n ) − 3 n 3 n ( n + 1 ) ( 2 n + 1 ) 6 = 6 − n ( n + 1 ) ( 2 n + 1 ) 2 n 3 {\displaystyle {\begin{aligned}R_{n}&=\sum _{i=1}^{n}(6-3({\frac {i}{n}})^{2})({\frac {1}{n}})\\&={\frac {6}{n}}\sum _{i=1}^{n}1-{\frac {3}{n^{3}}}\sum _{i=1}^{n}i^{2}\\&={\frac {6}{n}}(n)-{\frac {3}{n^{3}}}{\frac {n(n+1)(2n+1)}{6}}\\&=6-{\frac {n(n+1)(2n+1)}{2n^{3}}}\end{aligned}}}
So the integral is:
∫ 0 1 ( 6 − 3 x 2 ) d x = lim n → ∞ R n = lim n → ∞ ( 6 − n ( n + 1 ) ( 2 n + 1 ) 2 n 3 ) = 6 − lim n → ∞ 2 n 3 + 3 n 2 + n 2 n 3 = 6 − lim n → ∞ 2 + 3 / n + 1 / n 2 2 = 6 − 2 − 0 − 0 2 = 6 − 1 = 5 {\displaystyle {\begin{aligned}\int _{0}^{1}(6-3x^{2})dx&=\lim _{n\to \infty }R_{n}\\&=\lim _{n\to \infty }(6-{\frac {n(n+1)(2n+1)}{2n^{3}}})\\&=6-\lim _{n\to \infty }{\frac {2n^{3}+3n^{2}+n}{2n^{3}}}\\&=6-\lim _{n\to \infty }{\frac {2+3/n+1/n^{2}}{2}}\\&=6-{\frac {2-0-0}{2}}=6-1=5\end{aligned}}}
Checking our answer, we see that
∫ 0 1 ( 6 − 3 x 2 ) d x = ( 6 x − x 3 ) ∣ 0 1 = 6 ( 1 ) − ( 1 ) 3 − ( 6 ( 0 ) − ( 0 ) 3 ) = 6 − 1 = 5 {\displaystyle \int _{0}^{1}(6-3x^{2})dx=(6x-x^{3})\mid _{0}^{1}=6(1)-(1)^{3}-(6(0)-(0)^{3})=6-1=5}
which matches the above.
. The interval is ![{\displaystyle [0,1]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
so 
. In this case 
so the right Riemann sum is:
