구분구적법

$\displaystyle \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n} \frac{2}{n} \left( 1 + \frac{2 k}{n} \right) ^{4}}$에서 $f ( x ) = x ^{4}$, $\displaystyle x _{k} = 1 + \frac{2 k}{n}$라 하면 $\displaystyle \delta x = \frac{2}{n}$, $x _{0} = 1$, $x _{n} = 3$ 따라서 $\displaystyle \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n} \frac{2}{n} \left( 1 + \frac{2 k}{n} \right) ^{4}}$$\displaystyle = \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{n} f ( x _{k} ) \delta x}$ $\displaystyle = \int _{1} ^{3} {f ( x )} dx$$\displaystyle = \int _{1} ^{3} {x ^{4}} dx$$\displaystyle = \left[ \frac{1}{5} x ^{5} \right] _{1} ^{3}$$\displaystyle = \frac{1}{5} ( 3 ^{5} - 1 )$$\displaystyle = \frac{242}{5}$ 이므로 $5 a = 242$