Approximate each definite integral in Exercises 41–52 with the indicated method and to the given degree of accuracy. Then calculate the definite integral exactly, and verify that the error bounds you found are accurate.
${\int }_{-1}^{1}x{e}^{x}dx,\mathrm{left}\mathrm{sum}\mathrm{within}0.5$

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

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" $$\begin{gathered} ∆x=\frac{1-(-1)}{n} \\ =\frac{2}{n} \end{gathered}$$

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

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

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