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Using the definition of the definite integral as a limit of sums compute $f_{a}^{b} xdx$ for any a,b belonging to the set of real numbers where a<b.

And my answer is 1/2 which i don`t think is right.

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    I don't understand your question. Can you rewrite it? Are you asking how to solve $\int_a^b x dx$? If I'm right, so how did you get $1/2$? – Tio Zuca Nov 25 '23 at 16:14
  • @TioZuca Yes im solving that integral, well i don`t know how to type it up, but i used the definition of the definite integral as a limit of sums. – cosygod Nov 25 '23 at 16:24

1 Answers1

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The integral $\int_a^b xdx = \lim_{n\rightarrow \infty} \sum_{k=0}^n x_k \Delta x$, where $x_k=a+k\Delta x$, $\Delta x =\frac{b-a}{n}$. So $$\int_a^b x dx=\lim_{n\rightarrow \infty}\frac{b-a}{n}\sum_{k=0}^n \left( a+\frac{b-a}{n}k\right)=\lim_{n\rightarrow\infty}\frac{b-a}{n}\left( an + \frac{(b-a)(n+1)}{2} \right)=\lim_{n\rightarrow\infty} \left(a(b-a)+\frac{(b-a)^2}{2}\frac{n+1}{n}\right)=ab-a^2+\frac{b^2-2ab+a^2}{2}=\frac{b^2}{2}-\frac{a^2}{2}. \blacksquare$$

Tio Zuca
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