The area of the figure enclosed by y = ex, x = 0 and y = e is______ ．

The area of the figure enclosed by y = ex, x = 0 and y = e is______ ．

From the problem meaning y = E & nbsp; y = ex & nbsp;, it is found that the intersection coordinates are (1, e), so the area of the figure enclosed by the straight line y = e, Y axis and the curve y = ex is: ∫ 01 (e-ex) DX = (ex ex Ex) | 10 = 1

It is known that the parabola y = - 2x ^ 2-12x-20 is translated according to the vector a, so that its vertex is on the straight line x = 2 and the chord length cut on the X axis is 6?

Let a = (s, t), parabola y = - 2x & sup2; - 12x-20 translate according to vector a to be:
Y-T = - 2 (X-S) & sup2; - 12 (X-S) - 20, that is y = - 2x & sup2; + (4s-12) x + (12s-20 + T)
The vertex on the line x = 2 is 2 = (4s-12) / 2 × (- 2). S = 5. Y = - 2x & sup2; + 8x + 40 + t
The chord length cut on the x-axis is 6:2 √ [64 + 8 (40 + T)] / 4 = 6, t = - 30
A = (5, - 30), y = - 2x & sup2; + 8x + 10

Let p be any point on the parabola y = 2x ^ 2 + 1, point a (0, - 1), point m such that vector PM = 2, vector Ma, then the trajectory equation of M is

Let P (a, 2A & # 178; + 1), m (x, y), then vector PM = (x-a, y-2a & # 178; - 1), vector Ma = (- x, - 1-y)
∵ vector PM = 2 vector Ma
∴x-a =2(-x) ①
y-2a²-1=2( -1-y) ②
A = 3x is obtained from (1), and y = 6x & # 178; - 1 / 3 is obtained from (2)