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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.

Sagot :

LammettHash

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

[tex]\Delta x = \dfrac{3 - 1}4 = \dfrac12[/tex]

The left endpoint of the [tex]i[/tex]-th interval is

[tex]\ell_i = 1 + \dfrac i2[/tex]

The area under the curve is then approximately

[tex]\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}[/tex]

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