If f(x) = operatorname{Max}{x^3-4, x^4-4} and | vedclass

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(A) We need to evaluate $I = \int_{-1}^1 (f(x) - g(x)) \, dx = \int_{-1}^1 f(x) \, dx - \int_{-1}^1 g(x) \, dx$. First,consider $f(x) = \operatorname{

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$-\frac{151}{20}$

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(A) We need to evaluate $I = \int_{-1}^1 (f(x) - g(x)) \, dx = \int_{-1}^1 f(x) \, dx - \int_{-1}^1 g(x) \, dx$. First,consider $f(x) = \operatorname{Max}\{x^3-4, x^4-4\}$. Since $x^4 \ge x^3$ for $x \in [-1, 0]$ and $x^4 \le x^3$ for $x \in [0, 1]$,we have $f(x) = x^4-4$ for $x \in [-1, 0]$ and $f(x) = x^3-4$ for $x \in [0, 1]$. $\int_{-1}^1 f(x) \, dx = \int_{-1}^0 (x^4-4) \, dx + \int_0^1 (x^3-4) \, dx = [\frac{x^5}{5} - 4x]_{-1}^0 + [\frac{x^4}{4} - 4x]_0^1 = (0 - (-\frac{1}{5} + 4)) + (\frac{1}{4} - 4) = -\frac{19}{5} - \frac{15}{4} = -\frac{76+75}{20} = -\frac{151}{20}$. Next,consider $g(x) = \operatorname{Min}\{x^2, x^3\}$. For $x \in [-1, 0]$,$x^3 \le x^2$,so $g(x) = x^3$. For $x \in [0, 1]$,$x^2 \le x^3$,so $g(x) = x^2$. $\int_{-1}^1 g(x) \, dx = \int_{-1}^0 x^3 \, dx + \int_0^1 x^2 \, dx = [\frac{x^4}{4}]_{-1}^0 + [\frac{x^3}{3}]_0^1 = (0 - \frac{1}{4}) + (\frac{1}{3} - 0) = -\frac{1}{4} + \frac{1}{3} = \frac{1}{12}$. Finally,$I = -\frac{151}{20} - \frac{1}{12} = \frac{-453 - 5}{60} = -\frac{458}{60} = -\frac{229}{30}$. Wait,re-evaluating the options,let's re-check the integral calculation. $\int_{-1}^1 f(x) dx = -\frac{151}{20}$. $\int_{-1}^1 g(x) dx = \frac{1}{12}$. The result is $-\frac{229}{30}$. Given the options provided,there might be a typo in the question or options. However,based on the standard calculation,the result is $-\frac{229}{30}$.

If $f(x) = \operatorname{Max}\{x^3-4, x^4-4\}$ and $g(x) = \operatorname{Min}\{x^2, x^3\}$,then $\int_{-1}^1 (f(x) - g(x)) \, dx =$

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