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

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(B) To find the area bounded by the curve $y = x^2 - x$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y = 0$.$x

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(B) To find the area bounded by the curve $y = x^2 - x$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y = 0$.$x^2 - x = 0 \implies x(x - 1) = 0$.So,the intersection points are $x = 0$ and $x = 1$.The curve $y = x^2 - x$ lies below the $X$-axis in the interval $(0, 1)$ because for any $x \in (0, 1)$,$x^2 The area $A$ is given by the integral:$A = \left| \int_{0}^{1} (x^2 - x) \, dx \right|$$A = \left| \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} \right|$$A = \left| \left( \frac{1}{3} - \frac{1}{2} \right) - (0 - 0) \right|$$A = \left| -\frac{1}{6} \right| = \frac{1}{6}$ sq. units.

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

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