limn→∞∑i=1n1n⋅ei/n\displaystyle\lim_{n\to\infty}\sum_{i=1}^n \dfrac{1}{n}\cdot e^{i/n}n→∞limi=1∑nn1⋅ei/n equalsA111B000Ce−1e - 1e−1check_circleDeeeExplanationRight Riemann sum on [0,1][0, 1][0,1] for f(x)=exf(x) = e^xf(x)=ex. Limit =∫01ex dx=e−1= \int_0^1 e^x\,dx = e - 1=∫01exdx=e−1.