25. soit la fonction f(x)=x² + 2x définie sur lintervalle -1, 4. a) évaluez rₙ = ∑ₖ₌₁ⁿf(xₖ*)δx lorsque xₖ*…
25. soit la fonction f(x)=x² + 2x définie sur lintervalle -1, 4. a) évaluez rₙ = ∑ₖ₌₁ⁿf(xₖ*)δx lorsque xₖ* représente la borne supérieure du k - ième des n sous - intervalles de longueur δx formés par une partition régulière de lintervalle -1, 4.
Answer
Explanation:
Step1: Calculate $\Delta x$
The interval is $[a,b]=[-1,4]$ and we divide it into $n$ sub - intervals. Then $\Delta x=\frac{b - a}{n}=\frac{4-(-1)}{n}=\frac{5}{n}$.
Step2: Find the right - hand endpoints $x_{k}^*$
The right - hand endpoints of the sub - intervals for a regular partition of $[a,b]$ are given by $x_{k}^=a + k\Delta x$. Here, $a=-1$ and $\Delta x=\frac{5}{n}$, so $x_{k}^=-1+\frac{5k}{n}$ for $k = 1,2,\cdots,n$.
Step3: Calculate $f(x_{k}^*)$
We have $f(x)=x^{2}+2x$. Substitute $x_{k}^=-1+\frac{5k}{n}$ into $f(x)$: [ \begin{align} f(x_{k}^)&=\left(-1 + \frac{5k}{n}\right)^{2}+2\left(-1+\frac{5k}{n}\right)\ &=1-\frac{10k}{n}+\frac{25k^{2}}{n^{2}}-2+\frac{10k}{n}\ &=\frac{25k^{2}}{n^{2}}-1 \end{align} ]
Step4: Calculate $R_{n}=\sum_{k = 1}^{n}f(x_{k}^*)\Delta x$
[ \begin{align*} R_{n}&=\sum_{k = 1}^{n}\left(\frac{25k^{2}}{n^{2}}-1\right)\frac{5}{n}\ &=\sum_{k = 1}^{n}\left(\frac{125k^{2}}{n^{3}}-\frac{5}{n}\right)\ &=\frac{125}{n^{3}}\sum_{k = 1}^{n}k^{2}-\frac{5}{n}\sum_{k = 1}^{n}1 \end{align*} ] We know that $\sum_{k = 1}^{n}k^{2}=\frac{n(n + 1)(2n+1)}{6}$ and $\sum_{k = 1}^{n}1=n$. [ \begin{align*} R_{n}&=\frac{125}{n^{3}}\cdot\frac{n(n + 1)(2n + 1)}{6}-\frac{5}{n}\cdot n\ &=\frac{125}{6}\cdot\frac{(n^{2}+n)(2n + 1)}{n^{3}}-5\ &=\frac{125}{6}\cdot\frac{2n^{3}+3n^{2}+n}{n^{3}}-5\ &=\frac{125}{6}\left(2+\frac{3}{n}+\frac{1}{n^{2}}\right)-5\ &=\frac{125}{3}+\frac{125}{2n}+\frac{125}{6n^{2}}-5\ &=\frac{110}{3}+\frac{125}{2n}+\frac{125}{6n^{2}} \end{align*} ]
Answer:
$R_{n}=\frac{110}{3}+\frac{125}{2n}+\frac{125}{6n^{2}}$