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

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(D) The area $A$ is given by the integral of the absolute value of the function $y = x^2 - x - 6$ from $x = -1$ to $x = 1$.First,we check if the curve

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$\frac{34}{3}$

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(D) The area $A$ is given by the integral of the absolute value of the function $y = x^2 - x - 6$ from $x = -1$ to $x = 1$.First,we check if the curve crosses the $x$-axis in the interval $[-1, 1]$.Setting $y = 0$,we get $x^2 - x - 6 = 0$,which factors as $(x - 3)(x + 2) = 0$.The roots are $x = 3$ and $x = -2$.Since neither $3$ nor $-2$ lies in the interval $[-1, 1]$,the function $y = x^2 - x - 6$ does not change sign in this interval.For $x \in [-1, 1]$,$x^2 - x - 6$ is negative (e.g.,at $x = 0$,$y = -6$).Thus,the area is $A = \int_{-1}^{1} |x^2 - x - 6| \, dx = \int_{-1}^{1} -(x^2 - x - 6) \, dx$.$A = - \left[ \frac{x^3}{3} - \frac{x^2}{2} - 6x \right]_{-1}^{1}$.$A = - \left[ (\frac{1}{3} - \frac{1}{2} - 6) - (-\frac{1}{3} - \frac{1}{2} + 6) \right]$.$A = - \left[ \frac{1}{3} - \frac{1}{2} - 6 + \frac{1}{3} + \frac{1}{2} - 6 \right]$.$A = - \left[ \frac{2}{3} - 12 \right] = - \left[ \frac{2 - 36}{3} \right] = - \left[ -\frac{34}{3} \right] = \frac{34}{3}$ sq. units.

The area of the region bounded by the curve $y = x^2 - x - 6$,the $x$-axis $(y = 0)$,and the lines $x = -1$ and $x = 1$ is . . . . . . sq. units.

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