## Calculus: Early Transcendentals 8th Edition

The area for $G_{i}=i^{3}$ The area of ABCD is $\sum_{i=1}^ni^3=[\frac{n(n+1)}{2}]^2$
Area for $G_{i}=[\frac{i(i+1)}{2}]^{2}-[\frac{i(i-1)}{2}]^{2}$ $=\frac{i^{2}(i+1)^{2}}{4}-\frac{i^{2}(i-1)^{2}}{4}$ $=\frac{1}{4}i^{2}[(i+1)^{2}-(i-1)^{2}]$ $=\frac{1}{4}i^{2}[4i]$ Hence, the area for $G_{i}=i^{3}$ So the area of ABCD is $\sum_{i=1}^ni^3=[\frac{n(n+1)}{2}]^2$