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Use Riemann sums and a limit to compute the exact area under the curve. y = x^2 + 2 on [0, 1]
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???
@waterineyes can u help?
I think what they want you to do is just set up an integral to calculate the area underneath the function x^2 + 2 on the interval [0, 1] Using a riemann sum and a limit you could express the area as \[\large \lim_{n \rightarrow \infty} \sum_{i = 1}^n \mathrm{f}(x_i)\triangle x\] where \(\large \triangle x = \frac{1 - 0}{n}\) and \(\large x_i = 0 + i\triangle x\) Which is equal to \[\large \int_0^1 \mathrm{f}(x) dx = \int_0^1 x^2 + 2\,\,\, dx = \left[\frac{x^3}{3}+2x\right]^1_0\]
okay got you
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