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

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(C) To find the area bounded by the curve $y = x^2 - x - 6$ and the $X$-axis,we first find the points where the curve intersects the $X$-axis by setti

Vedclass · Accepted Answer

$\frac{125}{6}$

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(C) To find the area bounded by the curve $y = x^2 - x - 6$ and the $X$-axis,we first find the points where the curve intersects the $X$-axis by setting $y = 0$.$x^2 - x - 6 = 0$$(x - 3)(x + 2) = 0$So,the intersection points are $x = -2$ and $x = 3$.The area $A$ is given by the integral of the absolute value of the function from $x = -2$ to $x = 3$:$A = \int_{-2}^{3} |x^2 - x - 6| \, dx$Since the parabola opens upward and lies below the $X$-axis between its roots,the value of $y$ is negative in this interval.$A = - \int_{-2}^{3} (x^2 - x - 6) \, dx$$A = - [\frac{x^3}{3} - \frac{x^2}{2} - 6x]_{-2}^{3}$Evaluating the definite integral:At $x = 3$: $(\frac{27}{3} - \frac{9}{2} - 18) = 9 - 4.5 - 18 = -13.5$At $x = -2$: $(\frac{-8}{3} - \frac{4}{2} + 12) = -2.666 - 2 + 12 = 7.333$$A = - (-13.5 - 7.333) = - (-20.833) = 20.833$Converting to fraction: $20.833 = \frac{125}{6}$ sq. units.

The area of the region bounded by the curve $y = x^2 - x - 6$ and the $X$-axis is . . . . . . sq. units.

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