Application of Integration

S >

1

e

(As area of rectangle OCDS = 1/e)
Since w−x2 ≥ w−x for all x ∊ [0 , 1] ⇒ S >

1

∫

0

e−xdx = (1−

1

e

)
Area of rectangle OAPQ + Area of rectangle QBRS > S
S <

1

√2

(1)+(1−

1

√2

)(

1

√e

)
since

1

4

(1+

1

√e

) < 1−

1

e

Hence, option (C) is incorrect.