For help and elimination, x = (√ 3 / 4) (T1-T2), y = (1 / 4) (T1 + T2), T1 ^ 2 + t1t2 + T2 ^ 2 = 12

For help and elimination, x = (√ 3 / 4) (T1-T2), y = (1 / 4) (T1 + T2), T1 ^ 2 + t1t2 + T2 ^ 2 = 12

t1-t2=✓3／4x,t1^2-2t1t2+t2^2=3／16x^2t1+t2=1／4y,t1^2+2t1t2+t2^2=1／16y^2,12+t1t2=1／16y^24t1t2=1／16y^2-3／16x^2t1t2=1／64y^2-3／64x^212+1／64y^2-3／64x^2=1／16y^212-3／64x^2=3／64y^2(1／16x)^...

Higher integral ∫ (x ^ 2) * e ^ (x ^ 2) DX

A: it's very complicated
∫x^2e(x^2)dx
=(1/2)∫xe^(x^2)dx^2
=(1/2)∫xd(e^x^2)
=(1/2)xe^(x^2)-(1/2)∫e^x^2d(x^2)^(1/2)
=(1/2)xe^(x^2)-(1/4)∫e^x^2dx^2/(x^2)^(1/2)
∫e^x^2dx^2/(x^2)^(1/2)
Let t = (x ^ 2)
=∫e^tdt/t^(1/2)
e^t=1+t+t^2/2!+t^3/3!+..+t^n/n!
=∫[1/t^(1/2)+t^(1/2)+t^(3/2)/2!+t^(5/2)/3!+..+t^(n-1/2)/n!]dt
=2t^(1/2)+(2/3)t^(3/2)+(2/5)t^(5/2)/2!+(2/7)t^(7/2)/3!+..+(n+1/2)*t^(n+1/2)/n!+C
∫x^2e^(x^2)dx
=(1/2)xe^(x^2)-(1/4)[2*x+(2/3)x^3+(2/5)x^5/2!+(2/7)x^7/3!+..+(n+1/2)x^(2n+1)/n!] +C

If the higher number reaches the person's progression, find the integral (x ^ 3) * (e ^ - x ^ 4) DX,
After the college entrance examination, I have nothing to do to preview advanced mathematics, but I don't really know what's going on, why can I do it like this? What's going on? I kneel down to ask the real hero, take this question as an example,
What is the solution of X & # 179; DX = - D (- x ^ 4 / 4)?

∫x³(e^(-x^4)dx
=(-1/4)∫e^(-x^4)d(-x^4)
=(-1/4)*[e^(-x^4)]+C
Using integral: X & # 179; DX = - D (- x ^ 4 / 4)

In a triangle, it is the coordinate origin, a (1, cosx), B (SiNx, 1), where x is the first quadrant angle

In △ ABC, a (cosx, SiNx), B (1,0), C (0,1) x belong to (0, π / 2)
① Use X to represent the area of △ ABC
② Finding the maximum area of △ ABC

Thinking:
Because x belongs to (0, π / 2), so: cosx > 0, SiNx > 0, that is to say, point a is in the first quadrant, you can draw a sketch briefly
Then, the length of AC, AB and BC is calculated by using the distance formula between two points
Then we use the cosine theorem to find the cosine value of one angle, and then find the sine value of this angle
Then use the formula: S = 1 / 2absinc to get the answer to the first question
The first problem is solved, the second problem is simple, note: X belongs to (0, π / 2) this condition is OK

If the vertex of parabola y = x ^ 2 + X-6 is p, and its intersection with X axis is a and B, then the area of triangle PAB is______

125/8

If the vertex of parabola y = x2 + 4x is p, and the two intersections of parabola y = x2 + 4x and X axis are C and D, then the area of △ PCD is______ ．

∵ y = x2 + 4x = (x + 2) 2-4, ∵ P (- 2, - 4). Let y = 0, the equation x2 + 4x = 0, the solution, x = 0 or - 4, then CD = 4, then the area of △ PCD is 12 × 4 × 4 = 8. So the answer is 8

What is the area of the triangle bounded by the vertex of the parabola y = x-4x + 3 and its intersection with the x-axis?

X1 + x2 = 4, x1x2 = 3 | x1-x2 | = radical (16-12) = 2, vertex is (2, - 1) s = 2 * 1 / 2 = 1

The area of the triangle formed by the vertex of the parabola y = x2-4x + 3 and its intersection with the X axis is______ ．

From the meaning of the title, we can get: the ordinate of the vertex of the parabola is 4ac − b24a = - 1, the height of the bottom edge is 1; ∵ x2-4x + 3 = 0, the solution is X1 = 1, X2 = 3, the intersection of the parabola and the X axis is (1,0), (3,0); from the meaning of the title, we can get: the length of the bottom edge = | x1-x2 | = 2, ∵ the vertex of the parabola y = x2-4x + 3 and its intersection with the X axis, the triangle area enclosed by the three-point line is: 12 × 2 × 1 = 1

The image of the square of the parabola y = 3x moves two units to the right and one unit to the down. Its analytical expression is

The image of the square of y = 3x is shifted two units to the right to get y = 3 (X-2) ^ 2
Move down one more unit, and its analytical formula is y = 3 (X-2) ^ 2-1 = 3x ^ 2-12x + 11