91影视

The given integral is${鈭�}_{-1}^{1}x{e}^{x}dx,\mathrm{left}\mathrm{sum}\mathrm{within}0.5$

The derivative of the function is calculated below,

$f\text{'}\left(x\right)=x{e}^x+e^x$

The derivative is positive on the interval $[-1,1]$. Thus, the function is monotonically increasing on the given interval.

So, we get,

localid="1649403824529" role="math" $$\begin{gathered} 鈭�x=\frac{1-(-1)}{n} \\ =\frac{2}{n} \end{gathered}$$

And the magnitude of the error using n rectangular left sum has following bounds,

localid="1649403738872" $$\begin{gathered} \left|E_{LEFT\left(n\right)}\right|鈮�\left|f\left(b\right)-f\left(a\right)\right|鈭�x \\ \left|E_{LEFT\left(n\right)}\right|鈮�\left|1e^1-(-1)e^{-1}\right|\left(\frac{2}{n}\right) \\ \left|E_{LEFT\left(n\right)}\right|鈮�\frac{6.1723}{n} \end{gathered}$$

Now, to find the value of n such that $\left|E_{LEFT\left(n\right)}\right|\mathrm{is}\mathrm{less}\mathrm{than}0.5$,

$$\begin{gathered} \frac{6.1723}{n}<0.5 \\ \frac{6.1723}{0.5}<n \\ n>13 \end{gathered}$$

This means that $n=13$is the first positive integer for which the left sum will be within $0.5$of the actual area under the curve.

The left sum can be calculated as below,

localid="1649404917002" $$\begin{gathered} LEFT\left(13\right)=\underset{k=1}{\overset{13}{鈭�}}f\left(x_{k-1}\right)鈭�x \\ =(1e^1+2e^2+3e^3+4e^4+5e^5+6e^6+7e^7+8e^8+9e^9+10e^{10}+11e^{11}+12e^{12}+13e^{13})\left(\frac{1-(-1)}{13}\right) \\ =0.509077 \end{gathered}$$