Consider L = sqrt[3]{2012} + sqrt[3]{2013} + | vedclass

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(A) Let $f(x) = x^{1/3}$. Since $f''(x) = \frac{1}{3} \cdot \left(-\frac{2}{3}\right) x^{-5/3} = -\frac{2}{9} x^{-5/3}  0$,the function $

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$L + R < 2I$

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(A) Let $f(x) = x^{1/3}$. Since $f''(x) = \frac{1}{3} \cdot \left(-\frac{2}{3}\right) x^{-5/3} = -\frac{2}{9} x^{-5/3}  0$,the function $f(x)$ is strictly concave down.By the property of the definite integral for a concave function,the trapezoidal rule approximation $T$ with $n = 1000$ subintervals of width $h = 1$ satisfies $T The trapezoidal sum is $T = \frac{h}{2} [f(2012) + 2f(2013) + \ldots + 2f(3011) + f(3012)]$.Note that $L = \sum_{k=2012}^{3011} f(k)$ and $R = \sum_{k=2013}^{3012} f(k)$.Then $L + R = f(2012) + 2f(2013) + \ldots + 2f(3011) + f(3012) = 2T$.Since $T < I$,we have $2T < 2I$,which implies $L + R < 2I$.

Consider $L = \sqrt[3]{2012} + \sqrt[3]{2013} + \ldots + \sqrt[3]{3011}$,$R = \sqrt[3]{2013} + \sqrt[3]{2014} + \ldots + \sqrt[3]{3012}$,and $I = \int_{2012}^{3012} \sqrt[3]{x} \, dx$. Then,

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