Area bounded by the curves $y = \left\lbrack \frac{x^{2}}{64} + 2 \right\rbrack,\ y = x - 1$ and $x = 0$ above $x$-axis is ($\lbrack.\rbrack$ denotes the greatest integer function)

Question

Dr. Abubacker · Accepted Answer

(c)
$- 8 < x < 8\  \Rightarrow \ y = 2$
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$	herefore$ Required area $= \frac{1}{2}(1 + 3) 	imes 2$
$= 4\ sq\ unit$

Q. Area bounded by the curves $y = \left\lbrack \frac{x^{2}}{64} + 2 \right\rbrack,\ y = x - 1$ and $x = 0$ above $x$-axis is ($\lbrack.\rbrack$ denotes the greatest integer function)

Explanation

Reviewed by Dr. Abubacker

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