Consider region formed by the $y = 0, x = 2, y = 2$ .if the area enclosed by the curve and $y = I n x$ within this region, is being removed the area of the remaining region.

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Solution

Area formed by y = 0 , x = 2 , y = 2 = $2 * 2$ = $4$ sq units
Area being removed = $\int_{1}^{2} l n x d x$
$⟹ [x l n x - x]_{1}^{2}$
$⟹ 2 l n 2 - 2 + 1 = 2 l n 2 - 1$
Hence, area remaining = $4 - 2 l n 2 - (- 1)$ = $5 - 2 l n 2$ sq. units

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Consider region formed by the y=0,x=2,y=2.if the area enclosed by the curve and y=Inx within this region, is being removed the area of the remaining region.

Area formed by y = 0 , x = 2 , y = 2 = 2∗2 = 4 sq units Area being removed = ∫21lnxdx⟹[xlnx−x]21⟹2ln2−2+1=2ln2−1Hence, area remaining = 4−2ln2−(−1) = 5−2ln2 sq. units

Consider region formed by the $y = 0, x = 2, y = 2$ .if the area enclosed by the curve and $y = I n x$ within this region, is being removed the area of the remaining region.

Area formed by y = 0 , x = 2 , y = 2 = $2 * 2$ = $4$ sq units
Area being removed = $\int_{1}^{2} l n x d x$
$⟹ [x l n x - x]_{1}^{2}$
$⟹ 2 l n 2 - 2 + 1 = 2 l n 2 - 1$
Hence, area remaining = $4 - 2 l n 2 - (- 1)$ = $5 - 2 l n 2$ sq. units