If the area of the triangle formed by the line y = 2x + B and the two coordinate axes is 7, then B= The square of 3 (X-Y) is multiplied by the cube of [minus two thirds (Y-X)] and written in the form of the nth power of K (X-Y) A. The fifth power of minus 2 (X-Y) B. the sixth power of minus 2 (X-Y) C. The fifth power of minus 2 (X-Y) d. The sixth power of minus 2 (X-Y)

The intercept between the line and the x-axis is:
Let y = 0,2x + B = 0, x = B / 2
The intercept between the line and Y axis is:
Let x = 0, then y = B
Intercept are positive numbers, using absolute value to solve
(1 / 2) * absolute value of B / 2 * absolute value of B = 7
B²=28
B = plus or minus 2 √ 7

The area of the figure enclosed by the straight line y = 2x + B and the two coordinate axes is 9,

This graph may pass through quadrants 1,2,3 or 1,3,4,
The first case
1/2*(-b)*(-b/2)=9 b=6
The second situation
b=-6

The area of the triangle formed by the image and the coordinate axis of the linear function y = 2x + B is 4

Because the intersection point of the linear function y = 2x + B and the coordinate axis is: X-axis intersects a (- B / 2,0) and y-axis intersects P (0, b)
OA=｜-B/2｜OP=｜B｜
S=｜-B/2｜*｜B｜/2=4
B²=16
B = 4 or - 4
Y = 2x + 4 or y = 2X-4

The area of the common part of a graph bounded by r = 3cosx and R = 1 + cosx

This kind of problem is very troublesome, and the integral number is not easy to write
The first is the polar equation of the circle, and the second is the polar equation of the heart line
The first is transformed into a parametric equation: x = 3costcost; y = 3costsint
The second parameter equation is: x = (1 + cost) cost; y = (1 + cost) Sint
There are two intersections of the two curves, and the partial intersection of Y > 0 is t = π / 3
Only the area of the part y > 0. S = S1 + S2
=int(π/2,π/3)(3costsint*d(3cost^2))+int(π/3,0)((1+cost)sint*d((1+cost)cost))
Note that the part of S1 integral sign is: K1 = - 18cost ^ 2 * Sint ^ 2 * DT = (- 9 / 2) (1-cos2t ^ 2)
=(- 9 / 2) (1 / 2-cos4t / 2), so S1 = (- 9 / 2) int (π / 2, π / 3) (1 / 2-cos4t / 2) DT
=(-9/2)(t/2-sin4t/8)|(π/2,π/3)=3π/8-9sqrt(3)/32
Note that the part of S2 integral sign is: K2 = - (Sint + sintcost) * (Sint + 2sintcost) DT
=-(sint^2+3sint^2cost+2sint^2cost^2)dt
The first part: K21 = (cos2t-1) / 2 (DT), so S21 = sin2t / 4-T / 2 | (π / 3,0)
The second part: K22 = - 3sint ^ 2D (Sint), so S22 = - Sint ^ 3 | (π / 3,0)
The third part: K23 = (cos2t ^ 2-1) / 2 (DT) = (cos4t / 4-1 / 4) DT, so S23 = sin4t / 16-t / 4 | (π / 3,0)
So S2 = S21 + S22 + S23 = (π / 6-sqrt (3) / 8) + 3sqrt (3) / 8 + (sqrt (3) / 32 + π / 12)
=π/4+9sqrt(3)/32
So the area SS = 2S = 2 (S1 + S2) = 2 (3 π / 8-9sqrt (3) / 32 + π / 4 + 9sqrt (3) / 32) = 5 π / 4

Find the surface area of the surface with the center line r = a (1 + cosx) rotating around the polar axis

For the rest of the circle r = 3cos θ, from π / 3 integral to π / 2, S2 = 9 is still upper and lower symmetry, and the total area s = S1 + S2 = 3 π / 4-9 radical 3 / 8 + π / 2 + 9 radical 3 / 8 = 5 π / 4 θ

Area enclosed by curve
Figure area enclosed by y = x ^ 3 and y = x

The intersection coordinates of y = x ^ 3 and y = x are (0,0), (1,1) and (- 1, - 1)
Enclosed figure area s = 2 * ∫ (0,1) (x-x ^ 3) DX = 2 * [x ^ 2 / 2-x ^ 4 / 4] | (0,1)
=(1/2-1/4)*2
=1/2

In order to find out the area of the figure enclosed by the curve and the detailed explanation, the integral process is also necessary
Y = 1 / 2x ^ 2, x ^ 2 + y ^ 2 = 8 (both parts need to be calculated)

The enclosed figure is the upper and lower parts of the circle and parabola,
Left right symmetry, only the first quadrant,
The coordinates of the intersection point are (- 2,2) and (2,2),
S=2[∫(0→2) √（8-x^2)dx-∫(0→2) (x^2/2)dx]
=2[(0→2) (x/2)√（8-x^2)+(8/2)arcsin(x/2√2)-(0→2)x^3/6]
=2[2+π-4/3]
=4/3+2π.
Where √ (8-x ^ 2) integral, let x = 2 √ 2sint, DX = 2 √ 2costdt,
∫√（8-x^2)dx=2√2∫(cost)^2dt
=(x/2)√（8-x^2)+4arcsin(x/2√2).

If 2sin2 α = sin β + cos β, and sin ^ 2 α = sin β cos β, find the value of Cos2 α

Analysis: if 2sin2 α = sin β + cos β, then: 4sin & # 178; 2 α = (sin β + cos β) &# 178; = 1 + 2Sin β cos β and Sin & # 178; α = sin β cos β, so: 4sin & # 178; 2 α = 1 + 2Sin & # 178; α, then: 4 (1-cos & # 178; 2 α) = 2-cos2 α 4cos & # 178; 2 α + Cos2 α - 2 = 0cos &

How to find the area of irregular figure?

1. Split the irregular figure into several regular figures, and then use the known information to solve the problem. 2. Fill method, fill the irregular figure into a regular figure, and then subtract the area of the filled figure from the total area

Calculate the area of the irregular figure below

It can be divided and patched