Using the method of integration,find the area | vedclass

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(A) The curve $|x|+|y|=1$ represents a square with vertices at $A(0,1)$,$B(1,0)$,$C(0,-1)$,and $D(-1,0)$.The curve is symmetric about both the $x$-axi

Vedclass · Accepted Answer

$2 	ext{ sq. units}$

Vedclass · Answer

(A) The curve $|x|+|y|=1$ represents a square with vertices at $A(0,1)$,$B(1,0)$,$C(0,-1)$,and $D(-1,0)$.The curve is symmetric about both the $x$-axis and the $y$-axis.Therefore,the total area is $4$ times the area of the region in the first quadrant $(OBA)$.In the first quadrant,$x \ge 0$ and $y \ge 0$,so the equation becomes $x+y=1$,or $y=1-x$.$	ext{Area} = 4 \int_{0}^{1} y \, dx = 4 \int_{0}^{1} (1-x) \, dx$$= 4 \left[ x - \frac{x^2}{2} ight]_{0}^{1}$$= 4 \left( (1 - \frac{1}{2}) - (0 - 0) ight)$$= 4 \left( \frac{1}{2} ight) = 2 	ext{ sq. units}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

Using the method of integration,find the area bounded by the curve $|x|+|y|=1$.

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