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How to find the integral of [(x + 2) / x ^ 2 + 3x + 4] DX? Sorry, it should be [(x + 2) / (x ^ 2 + 3x + 4)] DX
Half of it
Original = 1 / 2 ∫ (2x + 3 + 1) / (x ^ 2 + 3x + 4) DX, see? Then split it into two parts and calculate separately
Integral range of ∫ [0,2] (x ^ 3-3x) DX
Wouldn't it be that simple?
∫(0,2) (x^3 - 3x) dx
= 1/4*x^4 - 3/2*x^2,(0,2)
= 1/4*2^4 - 3/2*2^2
= 4 - 6
= -2
∫e∧(-3x+1)dx
∫e∧(-3x+1)dx
=-1/3*∫e∧(-3x+1)d(-3x+1)
=-e^(-3x+1)/3+C
In the plane rectangular coordinate system, the straight line C1: x = 2T + 2ay = − t (t is the parameter), the curve C2: x = 2cos θ y = 2 + sin θ (θ is the parameter), if C1 and C2 have a common point, then the value of real number a is______ ．
From the straight line C1: x = 2T + 2ay = − t (t is the parameter), the parameter t is eliminated, and x = 2a-2y ① From the curve C2: x = 2cos θ y = 2 + sin θ (θ is a parameter), the parameter θ is eliminated, and X2 + 4 (Y-2) 2 = 4 ② Because C1 and C2 have common points, the above quadratic equation with respect to y has a real number solution, so △ ≥ 0, that is, 4 (a + 2) 2-4 × 2 × (A2 + 3) ≥ 0, a2-4a + 2 ≤ 0, 2 − 2 ≤ a ≤ 2 + 2
Coordinate system and parametric equation
It is known that the polar coordinate of curve C is p = 6sine, (E is the angle, P refers to the letter which is similar to P). The plane rectangular coordinate system is established by taking the pole as the origin of the plane rectangular coordinate system and the polar axis as the positive half axis of X. the parameter equation of line L is x = (√ 2) T-1, y = (√ 2) t / 2, (t is the parameter), then the chord length obtained by the intersection of line L and curve C is_______ .
Please write the answer, the most important thing is to be able to write the steps or methods
The main idea is to transform the polar coordinate equation and parameter equation into ordinary equation. By using ρ = √ X & sup2; + Y & sup2;, sin θ = Y / ρ, we can get the equation of circle as X & sup2; + (Y-3) & sup2; = 9. By eliminating the parameter t, we can get the ordinary equation of straight line x + 2Y + 1 = 0
In the polar coordinate system, if O is the pole and the line passes through the center of circle C: ρ = 22cos θ and is perpendicular to the line OC, then the polar coordinate equation of the line is___ ．
The Cartesian coordinate equation of circle C: ρ = 22cos θ is (X-2) 2 + y2 = 2, so the center of circle C is (2,0), the straight line passing through the center of circle and perpendicular to OC is x = 2, and the polar coordinate equation is ρ cos θ = 2. So the answer is: ρ cos θ = 2
4-4 coordinate system and parameter equation of high school mathematics elective course
First, we can know that the coordinates of the center of a circle (2cos θ, 2-2cos2 θ) are. Then according to the relationship between the coordinates, Cos2 θ = 2cos & # 178; θ - 1, we can get the trajectory of the center of a circle. 2-2cos2 θ = 2-4cos & # 178; θ + 2 = - 4cos & # 178; θ + 4 = - (2cos θ) &# 178; + 4, so if the center of a circle is (x, y), the trajectory is
Y = - X & # 178; + 4 this is the first small question of the first question. You can try again for yourself
In the first question of the second question, we first change the linear equation into a normal function about X and y, that is, ρ sin (θ - π / 4) = ρ sin θ cos π / 4 - ρ cos θ sin π / 4 = ycos π / 4-xsin π / 4 = 2 ^ (- 1 / 2) (Y-X) = M. then we can calculate m by making it equal to 3 according to the formula of the distance from the point to the straight line
The problem of coordinate system and parameter equation in high school mathematics
"If ρ＞ 0,0 ≤ θ ＜ 2 π is specified, then the points in the plane can be represented by the unique polar coordinates (ρ, θ) except for the poles; meanwhile, the points represented by the polar coordinates (ρ, θ) are also uniquely determined." why does the range of θ not include 2 π?
0 and 2 π are the same
So just take one
Otherwise, a point on the polar axis can be expressed in two ways
High school mathematics problems in coordinate system
1. The distance between two fixed points is 6, and the sum of the square of the distance from point m to the two vertices is 26. Find the trajectory of point M. 2. Given that point a is a fixed point, segment BC slides on fixed line L, known that lbcl = 4, and the distance from point a to line L is 3, find the trajectory equation of the outer center of triangle ABC. Please write the process in detail
1. Set the coordinates of two points as: a (0,0), B (6,0). The coordinates of m point are: m (x, y). Of course, they can also be set as: a (0,0), B (0,6). There are two kinds of equations. Then: the trajectory equation of M is: (x-0) ^ 2 + (y-0) ^ 2 + (X-6) ^ 2 + (y-0) ^ 2 = 26. (1) or: (x-0) ^ 2 + (y-0) ^ 2 + (x-0) ^ 2 + (y-0) ^ 2 = 26. (2) simplify (1) to get: 2x ^ 2 + 2Y ^ 2 - 12x + 36 = 26; X ^ 2 + y ^ 2 - 6x = - 5; (x ^ 2 - 6x + 9) + y ^ 2 = 4; (x - 3) ^ 2 + y ^ 2 = 2 ^ 2. This is the equation of a garden: the center of the garden is: (3,0); the radius is: 2. The simplified formula (2) is: 2x ^ 2 + 2Y ^ 2 - 12Y + 36 = 26; X ^ 2 + y ^ 2 - 6y = - 5; (y ^ 2 - 6y + 9) + x ^ 2 = 4; (Y - 3) ^ 2 + x ^ 2 = 2 ^ 2. This is the equation of a garden: the center of the garden is: (0,3); the radius is: 2
In the polar coordinate system, the line L perpendicular to the polar axis intersects the circle ρ = 4 at two points a and B. If | ab | = 4, the polar coordinate equation of the line L is______ ．
From the polar coordinate equation of the circle ρ = 4, we know that the radius of the circle is 4, and the chord length | ab | of the line L cut by the circle is 4. If the center of the circle is O, then ∠ AOB = 60 ° and the distance between the pole and the line L is d = 4cos30 ° = 23, so the polar coordinate equation of the line is ρ cos θ = 23. So the answer is: ρ cos θ = 23
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