The area of the region (in sq. units) enclose | vedclass

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(B) The curve is given by $y=x^3-19x+30$. Factoring the polynomial,we get $y=(x+5)(x-2)(x-3)$.The roots are $x=-5, 2, 3$. The curve is above the $x$-a

Vedclass · Accepted Answer

$\frac{517}{2}$

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(B) The curve is given by $y=x^3-19x+30$. Factoring the polynomial,we get $y=(x+5)(x-2)(x-3)$.The roots are $x=-5, 2, 3$. The curve is above the $x$-axis on the interval $[-5, 2]$ and below the $x$-axis on the interval $[2, 3]$.The total area $A$ is given by:$A = \int_{-5}^{2} (x^3-19x+30) dx + \left| \int_{2}^{3} (x^3-19x+30) dx ight|$$A = \int_{-5}^{2} (x^3-19x+30) dx - \int_{2}^{3} (x^3-19x+30) dx$Evaluating the integral $\int (x^3-19x+30) dx = \frac{x^4}{4} - \frac{19x^2}{2} + 30x + C$.For the first part: $\left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x ight]_{-5}^{2} = (4 - 38 + 60) - (\frac{625}{4} - \frac{475}{2} - 150) = 26 - (\frac{625-950-600}{4}) = 26 - (-\frac{925}{4}) = 26 + 231.25 = 257.25 = \frac{1029}{4}$.For the second part: $\left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x ight]_{2}^{3} = (\frac{81}{4} - \frac{171}{2} + 90) - (4 - 38 + 60) = (\frac{81-342+360}{4}) - 26 = \frac{99}{4} - 26 = \frac{99-104}{4} = -\frac{5}{4}$.Total Area $A = \frac{1029}{4} - (-\frac{5}{4}) = \frac{1034}{4} = \frac{517}{2} 	ext{ sq. units}$.

The area of the region (in sq. units) enclosed by the curve $y=x^3-19x+30$ and the $x$-axis is

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