Really simple integral problem

How did you come up with:

\(\displaystyle x_{i} = \frac{6}{n - 1}\)

If:
\(\displaystyle \Delta x = x_{i} - x_{i - 1} = \frac{5 - (-1)}{n} = \frac{6}{n} \quad \mbox{Length of subintervals}\)

then consider your partition:

\(\displaystyle x_{0} = -1 \quad < \quad x_{1} = -1 + \frac{6}{n} \quad < \quad x_{2} = -1 + \frac{2 \cdot 6}{n} \quad < \quad \mbox{...} \quad < \quad \underbrace{x_{i} = -1 + \frac{6i}{n}}_{} \quad < \quad \mbox{...} \quad < \quad x_{n} = -1 + \frac{6n}{n} = 5\)

So:
\(\displaystyle \lim_{n \to \infty} \sum^{n}_{i =1} f(x_{i}) \Delta x\)

\(\displaystyle = \lim_{n \to \infty} \sum^{n}_{i =1} (1 + 3x_{i}) \Delta x\)

\(\displaystyle = \lim_{n \to \infty} \sum^{n}_{i =1} \left(1 + 3\left(-1 + \frac{6i}{n}\right)\right) \cdot \frac{6}{n}\)
 
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