The area bounded by the curve $y = x,\ x$-axis and ordinates $x = - 1$ to $x = 2$, is

Question

Dr. Abubacker · Accepted Answer

(d)
Required area
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$\left| \int_{- 1}^{0}{x\ dx} \right| + \left| \int_{0}^{2}{x\ dx} \right|$
$= \left| \left\lbrack \frac{x^{2}}{2} \right\rbrack_{- 1}^{0} \right| + \left| \left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{2} \right|$
$= \left| - \frac{1}{2} \right| + |2|$
$= 2 + \frac{1}{2} = \frac{5}{2}$ sq unit

Q. The area bounded by the curve $y = x,\ x$-axis and ordinates $x = - 1$ to $x = 2$, is

Explanation

Reviewed by Dr. Abubacker

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