Let S be the area of the region enclosed by y=e^ -x² , y=0, x=0 and x=1 ; then

Step-by-step solution

The given curve y=e^ -x² Draw a rough sketch of curve at x=0, y=1 and at x=1, y=1 / e y=e^ -x² d y d x =-2 x e^ -x² Now, S = area exclosed by curve = XYCO and area of rectangle OCYL = 1 e Clearly S> 1 e a is true. For x [0,1] x² -x e^ -x² e^ -x x [0,1] ₀¹ e^ -x² d x> ₀¹ e^ -x d x=1- 1 e S>1- 1 e (b) is true. array Now S < area of rectangle XADO + area of rectangle ZDCN S < 1 2 1+ (1- 1 2 ) 1 e S < 1 2 + 1 e (1- 1 2 ) ( d ) is true Also as 1 4 (1+ 1 e ) < 1- 1 e (c) is incorrect.