Find the area of the figure enclosed by y = x ^ 2-4x and y = 3x

∫[(2x-x^2)-(2x^2-4x)]dx=∫(6x-3x^2)dx
=3x^2-x^3|2,0=3*2^2-2^3-0=4

The area of a is calculated by the area of a figure enclosed by the curve y = 4x-x ^ 2-3 and the tangent line at a (0, - 3) and B (3,0)
And the volume of this plane figure rotating around the x-axis. Note that it is calculated by using definite integral,

Derivation of curve: y '= - 2x + 4
Through a tangent: y + 3 = 4x, that is y = 4x-3
Tangent through B: y = (- 2 * 3 + 4) (x-3), that is, y = - 2x + 6
The intersection of two tangents is (3 / 2,3)
From the graph, we know that s = [definite integral of 4x-3 - (4x-x ^ 2-3) from 0 to 3 / 2] + [definite integral of - 2x + 6 - (4x-x ^ 2-3) from 3 / 2 to 3]
=[x ^ 3 / 3 difference between 3 / 2 and 0] + [x ^ 3 / 3-3x ^ 2 + 9x difference between 3 and 3 / 2] = 9 / 4
V = [definite integral of (4x-3 - (4x-x ^ 2-3)) * 2 π x * x from 0 to 3 / 2] + [definite integral of (- 2x + 6 - (4x-x ^ 2-3)) * 2 π x * x from 3 / 2 to 3]
Forget the rest

Find the area of the figure enclosed by the curves y = x ^ 2-2x + 5 and y = - x ^ 2 + 4x + 1

Let f (x) = (- x ^ 2 + 4x + 1) - (x ^ 2-2x + 5) = - 2x ^ 2 + 6x-4 = - 2 (x-1) (X-2)
If f (x) = 0, we can get X1 = 1, X2 = 2
So x ∈ [1,2]
G(x)=∮f(x)dx=(-2/3x^3+3x^2-4x)+C
So the area s = g (2) - G (1) = 1 / 3

The area of the closed figure enclosed by the curve y2 = 2x and the straight line y = - x + 4 is___ ．

From the curve y2 = 2x and the straight line y = - x + 4, the intersection points of the parabola and the straight line are (2,2) and (8, - 4). Select y as the integral variable, and write the curve equation as x = Y22 and x = 4-y. s = ∫ 2-4 [(4-y) - Y22] dy = (4y-y22-y36) | 2-4 = 30-12 = 18

The parabola y = ax * x + BX + C passes through the point a (- 1,0) and passes through the two intersections of the straight line y = x-3 and the coordinate axis as B and C, if the point m is on the parabola in the fourth quadrant
And OM is perpendicular to BC, the perpendicular foot is D, and the coordinate of point m is obtained

B (3,0), C (0, - 3) are obtained from the known data
The parabola is y = (x + 1) (x-3) = x ^ 2-2x-3
∵ OM is perpendicular to BC, and the analytical expression of BC is y = x-3
The analytic expression of OM is y = - x (x > 0)
Solving the equations y = x ^ 2-2x-3
y=-x(x>0)
Get the coordinates of M ((1 + √ 13) / 2, - (1 + √ 13) / 2)

Parabola y = the area bounded by SiNx and X axis in the first quadrant
If so, it should be the area of the interval [0, π]
There is no limit to the scope of the topic. Shouldn't it be infinite?

The title is a bit ambiguous
We should make it clear in [0, π], otherwise there will be some problems

What is the area of the parabola y = - x ^ 2 + 1, X axis and Y axis in the first quadrant?
RT

Let the area of the triangle bounded by the line L & # 8321;: y = KX + k-1 and the line L & # 8322;: y = (K + 1) x + K (k is a positive integer) and X axis be SK, then S & # 8321; + S & # 8322; + +S2006=?

The intersection of line L & # 8321;: y = KX + k-1 and X axis is: ((1-k) / K, 0)
The intersection of line L & # 8322;: y = (K + 1) x + K and X axis is: (- K / (K + 1), 0)
The intersection of line L & # 8321;: y = KX + k-1 and line L & # 8322;: y = (K + 1) x + k is: (- 1, - 1)

The calculation formula of curvature and radius of curvature and the meaning of symbols in the formula
Calculate the curvature and radius of curvature formula & nbsp; and the meaning of the symbols in the formula. (some are very strange to understand) & nbsp; needless to say how to deduce those, refine some. & nbsp; I just want the meaning of the formula and the symbols in the formula

Curvature radius is the reciprocal of curvature. The curvature formula is as follows: curvature k = y '' / [(1 + (y ') ^ 2) ^ (3 / 2)], where y', Y "are the first and second derivative of function y to x respectively. Parameter form: let R (T) = (x (T), y (T)), curvature k = (X'Y" - X "Y ') / ((x') ^ 2 + (y ') ^ 2) ^ (3 / 2)

Curvature formula of space curve
Curve r = (x (T), y (T), Z (T)), some places write curvature k = | R '× R "| / (| R' |) ^ (3 / 2), some places write C = kn, C = R" is curvature vector, curvature K = | R "|, normal vector n = C / | C |, why not? Which is right?

Both of them are right. For the parametric equation of a curve, a very general quantity t can be used as a parameter (such as the angle between the tangent line of the curve and the x-axis, etc.), or the arc length s can be used as a parameter. For the parametric equation with the arc length as a parameter, most of the quantities representing the characteristics of the curve have relatively simple formulas, just like the curvature k = | R '', which is

Find the area of the figure enclosed by y = x ^ 2-4x and y = 3x