I have to evaluate the integral as the limit of a Riemann sum $$\int_{-1}^{5}(1+3x)dx$$ Here it's what I did: $$\int_{-1}^{5}(1+3x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\Delta x$$ I know that $$\Delta x=\frac{6}{n}$$ $$f(x_i)=-2+\frac{18i}{n}$$ Therefore the Riemann sum is $$\lim_{n\to\infty}\sum_{i=1}^{n}(-2+\frac{18i}{n})(\frac{6}{n})=\lim_{n\to\infty}\sum_{i=1}^{n}\frac{-12}{n}+\sum_{i=1}^{n}\frac{108i}{n^2}=\lim_{n\to\infty}[\frac{-12}{n}+\frac{108}{n^2}\frac{(n+1)(n))}{2}]$$ Evaluating the limit the answer was 54 this is clearly wrong because the integral is equal to 42. What did I do wrong ?
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ $$\sum_{i=1}^{n}\frac{-12}{n}=-12\ne\frac{-12}{n}\qquad (\text{for}\ n\geqslant2)$$ $\endgroup$– DidFeb 18, 2018 at 15:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
It's just a computational mistake:$$\sum_{i=1}^nf\left(-1+6\frac in\right)\frac6n=\frac{54+42n}n.$$
answered Feb 18, 2018 at 15:39