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Use left-endpoint approximation to estimate the area under the curve of f(x)=x^3 on [1,3]. Use n = 4. Using the same function use midpoint approximation to estimate the area.

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Split up the interval [1, 3] into 4 intervals of equal length,


\Delta x = \frac{3 - 1}4 = \frac12

The left endpoint of the
i-th interval is


\ell_i = 1 + \frac i2

The area under the curve is then approximately


\displaystyle \int_1^3 f(x) \, dx \approx \sum_(i=1)^4 f(\ell_i)\Delta x \\\\ ~~~~~~~~ = \frac12 \sum_(i=1)^4 \left(1 + \frac i2\right)^3 \\\\ ~~~~~~~~ = \frac12 \sum_(i=1)^4 \left(1 + \frac{3i}2 + \frac{3i^2}4 + \frac{i^3}8\right) = \boxed{27}

User Siddharth Sonone
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