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Question

# Let S be the area of the region enclosed by y=e−x2,y=0,x=0 and x=1. Then

A
S14(1+1e)
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B
S11e
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C
S1e
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D
S12+1e(11e)
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Solution

## The correct option is D S≤1√2+1√e(1−1√e)x∈[0,1]⇒x2≤x⇒−x2≥−x⇒e−x2≥e−x S=1∫0e−x2dx≥1∫0e−xdx=[−e−x]10=1−1e i.e., S≥1−1e⇒ option (2) is correct. S is the area under the curve y=e−x2 Hence, S≥ area of the rectangle OPQS i.e., S≥1×1e=1e Hence, option (1) is correct. S=1∫0e−x2dx =1√2∫0e−x2dx+1∫1√2e−x2dx ≤ Area A+ Area B Now, Area A=1√2∫01.dx=1√2 Area B=1∫1√21√edx=1√e(1−1√2) Hence, S≤1√2+1√e(1−1√2) Hence, option (4) is correct Now, consider the integral e∫1dxx3/2=2[1−1√e] Substitute t=lnx ⇒e∫1dxx3/2=1∫0e−t/2dt Now, for 0<x<1 e−x2>e−x/2 ⇒1∫0e−x2dx>2[1−1√2] Now, 2[1−1√e]−14(1+1√e)=74−94√e =7√e−94√e>0 So, option (3) is not correct.

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