The area of the closed igure bounded by $x = - 1,\ x = 2$ and $y = \left\{ \begin{array}{r} - x^{2} + 2\ ,\ x \leq 1 \ 2x - 1\ \ ,\ \ x 1 \end{array} ight.\$ and the $x$-axis is

Question

The area of the closed igure bounded by $x = - 1,\ x = 2$ and $y = \left\{ \begin{array}{r} - x^{2} + 2\ ,\ x \leq 1 \ 2x - 1\ \ ,\ \ x > 1 \end{array} ight.\$ and the $x$-axis is

Dr. Abubacker · Accepted Answer

(a)
Required area
$A = \int_{- 1}^{1}{\left( - x^{2} + 2 \right)dx + \int_{1}^{2}{(2x - 1)dx}}$
$= \left\lbrack - \frac{x^{3}}{3} + 2x \right\rbrack_{- 1}^{1} + \left\lbrack x^{2} - x \right\rbrack_{1}^{2}$
$= \frac{10}{3} + 2 = \frac{16}{3}\ sq\ unit$

Q. The area of the closed igure bounded by $x = - 1,\ x = 2$ and $y = \left\{ \begin{array}{r} - x^{2} + 2\ ,\ x \leq 1 \\ 2x - 1\ \ ,\ \ x > 1 \end{array} \right.\$ and the $x$-axis is

Explanation

Reviewed by Dr. Abubacker

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