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\(f(x)=x^{2}\). Find \(\int\limits_{\overline{0}}^{a}\) and \(\int\limits_{n}^{\overline{0}}f\)
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A) \(\Large \frac{a}{3},\frac{a}{3}\) |
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B) \(\Large \frac{a^{2}}{3},\frac{a^{2}}{3}\) |
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C) \(\Large \frac{a^{3}}{3},\frac{a^{3}}{3}\) |
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D) none of these |
Correct Answer:
| C) \(\Large \frac{a^{3}}{3},\frac{a^{3}}{3}\) |
Description for Correct answer:
Let \(\Large P\{ \frac{ra}{n};r=0,1,2,...n \}\)
be a partition on \([0,a].\) Then
\(\Large m_{r}=\frac{(r-1)^{2}a^{2}}{n^{2}}\)
\(\Large M_{r}=\frac{r^{2}a^{2}}{n^{2}};\ \Delta x_{r}=\frac{a}{n}\)
\(L(P,f)=\sum\limits_{r=1}^{n}m_{r}\Delta x_{r}\)
\(\Large =\sum\limits_{r=1}^{n}\frac{(r-1)^{2}a^{2}}{n^{2}}.\frac{a}{n}\)
\(\Large =\frac{a^{3}}{n^{3}}\sum\limits_{r=1}^{n}(r-1)^{2}\)
\(\Large =\frac{a^{3}}{n^{3}}\left(\frac{(n-1)n(2n-1)}{6}\right) \)
\(\Large =\frac{a^{3}}{6} \left( \left(1-\frac{1}{n}\right) \left(2-\frac{1}{n}\right) \right) \)
\(\Large U(P,f)=\sum\limits_{r=1}^{n}M_{r}\Delta x_{r}\)
\(\Large =\sum\limits_{r=1}^{n}\frac{r^{2}a^{2}}{n^{2}}=\frac{a^{3}}{n^{3}}\sum\limits_{r=1}^{n}r^{2}\)
\(\Large =\frac{a^{3}}{n^{3}}\frac{n(n+1)(2n+1)}{6}\)
\(=\frac{a^{3}}{6} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right)\)
Therefore
\(\Large \int\limits_{\overline{0}}^{a}f=\lim\limits_{n\rightarrow \infty}\frac{a^{3}}{6} \left(1-\frac{1}{n}\right) \left(2-\frac{1}{n}\right)=\frac{a^{3}}{3}\)
\(\Large \int\limits_{0}^{\overline{a}}f=\lim\limits_{n\rightarrow \infty}\frac{a^{3}}{6} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right)=\frac{a^{3}}{3}\)
Also,
\(\int\limits_{\overline{0}}^{a}f=\int\limits_{0}^{\overline{a}}f\Rightarrow f\in R[0,a]\)
i.e., f is Reimainn integrable (R-integrable).
Let \(\Large P\{ \frac{ra}{n};r=0,1,2,...n \}\)
be a partition on \([0,a].\) Then
\(\Large m_{r}=\frac{(r-1)^{2}a^{2}}{n^{2}}\)
\(\Large M_{r}=\frac{r^{2}a^{2}}{n^{2}};\ \Delta x_{r}=\frac{a}{n}\)
\(L(P,f)=\sum\limits_{r=1}^{n}m_{r}\Delta x_{r}\)
\(\Large =\sum\limits_{r=1}^{n}\frac{(r-1)^{2}a^{2}}{n^{2}}.\frac{a}{n}\)
\(\Large =\frac{a^{3}}{n^{3}}\sum\limits_{r=1}^{n}(r-1)^{2}\)
\(\Large =\frac{a^{3}}{n^{3}}\left(\frac{(n-1)n(2n-1)}{6}\right) \)
\(\Large =\frac{a^{3}}{6} \left( \left(1-\frac{1}{n}\right) \left(2-\frac{1}{n}\right) \right) \)
\(\Large U(P,f)=\sum\limits_{r=1}^{n}M_{r}\Delta x_{r}\)
\(\Large =\sum\limits_{r=1}^{n}\frac{r^{2}a^{2}}{n^{2}}=\frac{a^{3}}{n^{3}}\sum\limits_{r=1}^{n}r^{2}\)
\(\Large =\frac{a^{3}}{n^{3}}\frac{n(n+1)(2n+1)}{6}\)
\(=\frac{a^{3}}{6} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right)\)
Therefore
\(\Large \int\limits_{\overline{0}}^{a}f=\lim\limits_{n\rightarrow \infty}\frac{a^{3}}{6} \left(1-\frac{1}{n}\right) \left(2-\frac{1}{n}\right)=\frac{a^{3}}{3}\)
\(\Large \int\limits_{0}^{\overline{a}}f=\lim\limits_{n\rightarrow \infty}\frac{a^{3}}{6} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right)=\frac{a^{3}}{3}\)
Also,
\(\int\limits_{\overline{0}}^{a}f=\int\limits_{0}^{\overline{a}}f\Rightarrow f\in R[0,a]\)
i.e., f is Reimainn integrable (R-integrable).
Part of solved Real Analysis questions and answers : >> Elementary Mathematics >> Real Analysis
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