Theorem 4 says that if f is integrable on [a,b], then

$\displaystyle \int_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^nf(x_i)\Delta{x}$

where $\displaystyle \Delta{x}=\frac{b-a}{n}$ and $\displaystyle x_i=a+i\Delta{x}$

My question is $\displaystyle \int_{-1}^{5} (1+3x)dx$

I have $\displaystyle \Delta{x}=\frac{6}{n}$ ; $\displaystyle x_i=\frac{6i}{n}$

so $\displaystyle \int_{-1}^{5} (1+3x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^n\left[1+\frac{18i}{n}\right]\cdot\frac{6}{n}=$

$\displaystyle =\lim_{n \rightarrow \infty}\frac{6}{n}\left[\sum_{i=1}^n1+\frac{18}{n}\sum_{i=1}^ni\right]=$

$\displaystyle =\lim_{n \rightarrow \infty}\frac{6}{n}\left[n+\frac{18}{n}\cdot\frac{n(n+1)}{2}\right]=$

$\displaystyle =\lim_{n \rightarrow \infty}6+54\cdot\frac{n^2+n}{n^2}=60$

but by taking the antiderivative I get 42.