Finding indefinite integral ∫ 1 / √ (9-4X ^ 2) DX=

Finding indefinite integral ∫ 1 / √ (9-4X ^ 2) DX=

Let u = 2x,
∫1/√(9-4x^2)dx = (1/2) ∫1/√(9- u^2) du
= (1/2) * arcsin(u/3) + C
= (1/2) * arcsin( 2x/3) + C

Seeking integral ∫ (1 / 9 + 4x ^ 2) DX

The original formula = ∫ 1 / 9 [1 / (1 + 4x & # / 9)] DX
=∫1/9*3/2[1/[1+(2x/3)²]d(2x/3)
=1/6*arctan(2x/3)+C

Find ∫ (x-1) / (9-4X ^ 2) DX

Divided into 2 points:
∫ (x-1)/ （9-4x^2）dx
=(1/12)∫ 1/(3-2x)dx -∫ 5/ (3+2x）dx
=(1/24)(-ln|3-2x|-5ln|3+2x|)+C

∫x^2/(9+4x^2)dx

∫x^2/(9+4x^2)dx =1/4* ∫4x^2/(9+4x^2)dx
=1/4 * ∫[1-9/(9+4x^2)]dx
=1/4 *∫dx -1/4 * ∫[ 1/(1+(2x/3)^2)]dx
=1/4 *∫dx -1/4 * 3/2* ∫[ 1/(1+(2x/3)^2)]d(2x/3)
=x/4 - 3/8 * arctan(2x/3) + c

The vertex coordinates of the image of the function y = - X & # 178; + 2x + 3 are

y
= -x² + 2x + 3
= -x² + 2x - 1 + 4
= -(x - 1)² + 4
So vertex coordinates (1,4)

After moving the vertex coordinates of quadratic function y = - 2x to (- 3,2), the analytic expression of the function is——

The vertex of y = - 2x2 is (0,0), and the analytic expression of the function when it is moved to (- 3,2) is y = - 2 (x + 3) 2 + 2, that is, y = - 2x2-6x-16

It is known that the Quasilinear equation of parabola C1: y ^ 2 = 2px is x = - 2. The center of hyperbola C2 is at the origin, the axis of symmetry is the coordinate axis, and the focus of parabola C1 is a focus. Find the value of real number P. if hyperbola C2 passes through point P (root 2, root 3), find the standard equation of hyperbola C2

The axis of symmetry of C1 (set as a straight line x = n,) must have n-3 = 3-5 / 4. In this way, the axis of symmetry of C1 can be obtained. Further observation shows that the intersection of C2 and Y axis is (0,2), then there must be a point on C1 and this point

The opening direction of a parabola, the axis of symmetry is the same as y = 0.5x2, the ordinate of the vertex is - 1, and the parabola passes through the point (1,1)

Because the axis of symmetry is the same as y = 0.5 * x ^ 2, the axis of symmetry is y, the abscissa of the vertex is 0, and the ordinate of the vertex is - 1, so the vertex coordinates are (0, - 1), so the parabola can be written as y = a * (x-0) ^ 2-1, and the parabola passes through the point (1,1), so a = 2, so the functional relation of this parabola is y = 2 * x ^ 2-1

It is known that the vertex of the parabola is (2,3) and passes through the point (3,1)

It is known that the vertex of the parabola is (2,3),
It can be set as y = a (X-2) ² + 3;
And passing through point (3,1)
So a (3-2) ² + 3 = 1;
a+3=1;
a=-2;
The parabola is y = - 2 (X-2) &# 178; + 3;
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Given that the parabola passes through points a (- 5,0), B (1,0) and the ordinate of the vertex is 9 / 2, the analytic expression of quadratic function is obtained

The parabola passes through points a (- 5,0), B (1,0)
Let the analytic expression of quadratic function be y = a (x + 5) (x-1)
Abscissa of vertex = (- 5 + 1) △ 2 = - 2
Substituting (- 2,9 / 2) into the function equation, we get the following results:
-9a=9/2
a=-1/2
So: the analytic expression of quadratic function is: y = (- 1 / 2) (x + 5) (x-1)
That is y = - (1 / 2) x & # 178; - 2x + 5 / 2