Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 R (b) The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in part (a). 13 + 23 + 33 + + n3 = n(n + 1) 2 2

The summand (R?) is missing, but we can always come up with another one.Divide the interval [0, 1] into [tex]n[/tex] subintervals of equal length [tex]\dfrac{1-0}n=\dfrac1n[/tex]:[tex][0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right][/tex]Let's consider a left-endpoint sum, so that we take values of [tex]f(\ell_i)={\ell_i}^3[/tex] where [tex]\ell_i[/tex] is given by the sequence[tex]\ell_i=\dfrac{i-1}n[/tex]with [tex]1\le i\le n[/tex]. Then the definite integral is equal to the Riemann sum[tex]\displaystyle\int_0^1x^3\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i-1}n\right)^3\frac{1-0}n[/tex][tex]=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=1}^n(i-1)^3[/tex][tex]=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=0}^{n-1}i^3[/tex][tex]=\displaystyle\lim_{n\to\infty}\frac{n^2(n-1)^2}{4n^4}=\boxed{\frac14}[/tex]