The area of the region bounded by the curve y | vedclass

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(C) The area is given by the integral $A = \int_0^3 |x^3 - 4x^2 + 3x| dx$.First,factor the expression: $x^3 - 4x^2 + 3x = x(x^2 - 4x + 3) = x(x-1)(x-3

Vedclass · Accepted Answer

$\frac{37}{12}$

Vedclass · Answer

(C) The area is given by the integral $A = \int_0^3 |x^3 - 4x^2 + 3x| dx$.First,factor the expression: $x^3 - 4x^2 + 3x = x(x^2 - 4x + 3) = x(x-1)(x-3)$.The expression $x(x-1)(x-3)$ is positive in $(0, 1)$ and negative in $(1, 3)$.Thus,$A = \int_0^1 (x^3 - 4x^2 + 3x) dx - \int_1^3 (x^3 - 4x^2 + 3x) dx$.Evaluating the first integral: $\left[ \frac{x^4}{4} - \frac{4x^3}{3} + \frac{3x^2}{2} \right]_0^1 = \frac{1}{4} - \frac{4}{3} + \frac{3}{2} = \frac{3 - 16 + 18}{12} = \frac{5}{12}$.Evaluating the second integral: $\left[ \frac{x^4}{4} - \frac{4x^3}{3} + \frac{3x^2}{2} \right]_1^3 = \left( \frac{81}{4} - \frac{108}{3} + \frac{27}{2} \right) - \left( \frac{5}{12} \right) = \left( \frac{81}{4} - 36 + \frac{54}{4} \right) - \frac{5}{12} = \left( \frac{135}{4} - 36 \right) - \frac{5}{12} = \left( \frac{135 - 144}{4} \right) - \frac{5}{12} = -\frac{9}{4} - \frac{5}{12} = -\frac{27}{12} - \frac{5}{12} = -\frac{32}{12} = -\frac{8}{3}$.Since the second part is negative,we take the absolute value: $|-\frac{8}{3}| = \frac{8}{3}$.Total Area = $\frac{5}{12} + \frac{8}{3} = \frac{5 + 32}{12} = \frac{37}{12}$.

The area of the region bounded by the curve $y = |x^3 - 4x^2 + 3x|$ and the $X$-axis,for $0 \leq x \leq 3$,is

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