Area of region bounded by x-0- y-0 x-2- y-2- y-ex y- lnx is

The correct option is B 6 4ln2 Area=∫ _ 0 ^ 2 #10;( x ln x #160; ) dx +∫ _ 0 ^ 2 ( x ln y #10; #160; ) dy Area=[ x ^ 2 2 ( xln ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

Area of regio...

Question

Area of region bounded by $x = 0$ , $y = 0$ $x = 2$ , $y = 2$ , $y \leq e^{x}$ & $y \geq l n x$ is

6 - 4 l n 2

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4 l n 2 - 2

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2 l n 2 - 4

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6 - 2 l n 2

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Solution

The correct option is B $6 - 4 l n 2$
$A r e a = \int_{0}^{2} (x - ln x) d x + \int_{0}^{2} (x - ln y) d y A r e a = {[\frac{x^{2}}{2} - (x ln x - x)]}_{0}^{2} + {[\frac{y^{2}}{2} - (y ln y - y)]}_{0}^{2} A r e a = 4 - 4 ln 2 + 2 A r e a = 6 - 4 ln 2$

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Area of region bounded by x=0,y=0 x=2,y=2,y≤ex&y≥lnx is

The correct option is B 6−4ln2Area=∫20(x−lnx)dx+∫20(x−lny)dyArea=[x22−(xlnx−x)]20+[y22−(ylny−y)]20Area=4−4ln2+2Area=6−4ln2

Area of region bounded by $x = 0$ , $y = 0$ $x = 2$ , $y = 2$ , $y \leq e^{x}$ & $y \geq l n x$ is

The correct option is B $6 - 4 l n 2$
$A r e a = \int_{0}^{2} (x - ln x) d x + \int_{0}^{2} (x - ln y) d y A r e a = {[\frac{x^{2}}{2} - (x ln x - x)]}_{0}^{2} + {[\frac{y^{2}}{2} - (y ln y - y)]}_{0}^{2} A r e a = 4 - 4 ln 2 + 2 A r e a = 6 - 4 ln 2$