Let f(x) = x^3 - 3x^2 + 3x + 1 and g be the i | vedclass

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(B) Given $f(x) = x^3 - 3x^2 + 3x + 1 = (x-1)^3 + 2$.Let $y = g(x)$,then $x = f(y) = (y-1)^3 + 2$.We need to find the area bounded by $y = g(x)$ and t

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$\frac{1}{4}$

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(B) Given $f(x) = x^3 - 3x^2 + 3x + 1 = (x-1)^3 + 2$.Let $y = g(x)$,then $x = f(y) = (y-1)^3 + 2$.We need to find the area bounded by $y = g(x)$ and the $x$-axis from $x = 1$ to $x = 2$.This is equivalent to the area bounded by $x = f(y)$ and the $y$-axis from $y = g(1)$ to $y = g(2)$.Since $f(0) = 1$,$g(1) = 0$. Since $f(1) = 2$,$g(2) = 1$.Thus,the area $A = \int_0^1 f(y) dy = \int_0^1 ((y-1)^3 + 2) dy$.Let $u = y-1$,then $du = dy$. When $y=0, u=-1$; when $y=1, u=0$.$A = \int_{-1}^0 (u^3 + 2) du = [\frac{u^4}{4} + 2u]_{-1}^0 = (0) - (\frac{1}{4} - 2) = -\frac{1}{4} + 2 = \frac{7}{4}$.Wait,re-evaluating the integral: $A = \int_0^1 (y^3 - 3y^2 + 3y + 1) dy = [\frac{y^4}{4} - y^3 + \frac{3y^2}{2} + y]_0^1 = \frac{1}{4} - 1 + \frac{3}{2} + 1 = \frac{1}{4} + \frac{6}{4} = \frac{7}{4}$.Re-checking the question: The area bounded by $y=g(x)$ and $x$-axis from $x=1$ to $x=2$ is $\int_1^2 g(x) dx$. Let $x=f(y)$,$dx=f'(y)dy$. When $x=1, y=0$; when $x=2, y=1$.$\int_1^2 g(x) dx = \int_0^1 y f'(y) dy = [y f(y)]_0^1 - \int_0^1 f(y) dy = (1 \cdot f(1) - 0 \cdot f(0)) - \int_0^1 (y^3 - 3y^2 + 3y + 1) dy = 2 - \frac{7}{4} = \frac{1}{4}$.

Let $f(x) = x^3 - 3x^2 + 3x + 1$ and $g$ be the inverse of $f$. The area bounded by the curve $y = g(x)$ with the $x$-axis between $x = 1$ and $x = 2$ is (in square units):

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