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(D) Given $f(x) = 3x^3 + 2x$. We need to find the area bounded by $g(x)$,the $x-$axis,and the line $x = 5$. Since $g(x)$ is the inverse of $f(x)$,the

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$\frac{13}{4}$

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(D) Given $f(x) = 3x^3 + 2x$. We need to find the area bounded by $g(x)$,the $x-$axis,and the line $x = 5$. Since $g(x)$ is the inverse of $f(x)$,the area bounded by $g(x)$ from $x=0$ to $x=5$ is equivalent to the area bounded by $f(x)$ with the $y-$axis from $y=0$ to $y=5$. First,find the value of $x$ for which $f(x) = 5$: $3x^3 + 2x = 5 \implies 3x^3 + 2x - 5 = 0$. By inspection,$x = 1$ is a root because $3(1)^3 + 2(1) - 5 = 0$. The area is given by $\int_{0}^{5} g(x) \, dx$. Using the property of inverse functions,$\int_{0}^{a} g(x) \, dx + \int_{0}^{g(a)} f(x) \, dx = a \cdot g(a)$. Here $a = 5$ and $g(5) = 1$. So,$\int_{0}^{5} g(x) \, dx + \int_{0}^{1} (3x^3 + 2x) \, dx = 5 \cdot 1 = 5$. $\int_{0}^{1} (3x^3 + 2x) \, dx = [\frac{3x^4}{4} + x^2]_{0}^{1} = \frac{3}{4} + 1 = \frac{7}{4}$. Therefore,the required area is $5 - \frac{7}{4} = \frac{20 - 7}{4} = \frac{13}{4}$.

Let $y = g(x)$ be the inverse of a bijective mapping $f : R \rightarrow R$ defined by $f(x) = 3x^3 + 2x$. The area bounded by the graph of $g(x)$,the $x-$axis,and the ordinate at $x = 5$ is:

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