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

S>1/e (As area of rectangle OCDS=1/e)
Since e-x²≥e-x ∀x∈[0,1].
⇒S>∫01e-xdx=(1−1/e)
Area of rectangle OAPQ + Area of rectangle QBRS > S
⇒S<1/√2(1)+(1−√2)(√e)
Since 1/4(1+√e)<1−√e