Use Reimann's limit process to find the area of the region between the graph of the function and the x-axis over the indicated interval. y=3x-4 Interval: [2,5]
Area of the curve y = 3x - 4 on the interval [2,5].\[A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f (c_i) \Delta x\]\[\Delta x = \frac {b-a}{n} = \frac {5 - 2}{n} = \frac {3}{n}\]\[c_i = a + i \Delta x = 2 + \frac {3i}{n}\]\[A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f (2 + \frac {3i}{n})(\frac {3}{n})\]\[A = \lim_{n \rightarrow \infty} \sum_{i =1}^{n} \left[ 3(2 + \frac {3i}{n}) - 4 \right] (\frac {3}{n})\]\[A = \lim_{n \rightarrow \infty} \sum_{i =1}^{n} \left[ \frac {6}{n} + \frac {27i}{n^2} \right]\]\[A = \lim_{n \rightarrow \infty} \frac {6}{n} \sum_{i =1}^{n} 1 + \frac{27}{n^2} \sum_{i = 1}^{n}i\]Applying the infinite limit, that simplifies down to...\[A = \frac {6}{n} * \frac {n}{1} + \frac{27}{n^2} * \frac {n^2}{2}\]\[A = 6 + \frac {27}{2} = 6 + 13.5 = 19.5\]
I haven't used this method in a while, thanks for the practice! :D
thanks
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