The area of the figure enclosed by the curve y = x2-1 and the straight line x + y = 1 is______ ．

The area of the figure enclosed by the curve y = x2-1 and the straight line x + y = 1 is______ ．

From y = x2-1 and the line x + y = 1, the intersection points are (- 2,3) and (1,0). Therefore, the area of the closed graph enclosed by y = x2-1 and the line x + y = 1 is s = ∫ 1 − 2 (1-x-x2 + 1) DX = (2x-12x2-13x3) | 1 − 2 = 92

Find the area of closed figure enclosed by curve y ^ 2 = x and straight line x = 1
Come on!
Do it with points!

S=∫(-1,1) （1-y²）dy
=∫(-1,1) 1dy-∫(-1,1) y²dy
=2-2/3
=4/3

Limx * LNX (x tends to positive zero)

=lim lnx/(1/x)
From the law of lobida
=lim (1/x)/(-1/x^2)
=-lim x
=0

Limx tends to 1 LNX / X-1
How to solve the problem

Did you learn the law of lobida
Type 0 / 0
Up and down derivation
=lim(1/x)/1
=lim(1/x)
=1