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.
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.
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 $$\int_a^b x dx=\lim_{n\rightarrow \infty}\frac{b-a}{n}\sum_{k=0}^n \left( a+\frac{b-a}{n}k\right)=\lim_{n\rightarrow\infty}\frac{b-a}{n}\left( an + \frac{(b-a)(n+1)}{2} \right)=\lim_{n\rightarrow\infty} \left(a(b-a)+\frac{(b-a)^2}{2}\frac{n+1}{n}\right)=ab-a^2+\frac{b^2-2ab+a^2}{2}=\frac{b^2}{2}-\frac{a^2}{2}. \blacksquare$$