Find the area of the plane figure enclosed by the square of the curve y = 4-x, y = the square of X - 2x Specific process, thank you!

Find the area of the plane figure enclosed by the square of the curve y = 4-x, y = the square of X - 2x Specific process, thank you!

Let y = 4-x ^ 2, y '= x ^ 2-2x, f (x) = Y-Y'
Let y = y '
The intersection coordinates of the two equations are (2,0) and (- 1,3)
So the area is: the integral of F (x) from - 1 ~ 2
Because f (x) = 4-2x ^ 2 + 2x
Therefore, the inverse derivation of F (x) gives f (x) = 4x - (2 / 3) x ^ 2 + x ^ 2
Therefore, the area is f (2) - f (- 1) (substituting - 1 and 2 into the expression of F (x) and subtracting them respectively is the basic use of calculus. It is difficult to do it at the beginning, so it is good to get used to it.)

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

Replace y = x + 4 with y = 1 / 2x ^ 2
1/2x^2=x+4
x^2-2x-8=0
(x-4)(x+2)=0
x=4 x=-2
So s area=
∫(-2->4)(x+4 -1/2x^2)dx
=∫(-2->4)xdx+4∫(-2->4)dx-1/2∫(-2->4)x^2dx
=1/2x^2|(-2->4)+4x|(-2->4)-1/6 x^3|(-2->4)
=1/2(4^2-2^2)+4(4-(-2))-1/6 * (4^3-(-2)^3)
=6+4*6-1/6*(64+8)
=6+24-12
=18

Find the area of the plane figure enclosed by the curve y = LNX and the straight line x = E and X axis

The area of the circumference x is integral from 1 to E
So the definite integral ∫ [1, e] lnxdx
=xlnx[1,e]-∫[1,e]dx
=e-(e-1)
=1
So the area is 1

The area of the plane figure surrounded by the curve y = LNX, the straight line x = 1, x = E and the X axis, and the volume of the rotating body generated by rotating around the X axis and Y axis respectively Thank you very much for your speed Please.

1) ∫lnxdx=[xlnx-x]|=1.
2) Around X axis
V1=∫πy²dx
=π∫ln²xdx
=π[xln²x]|-π∫2lnxdx
=π(e-2).
3) Around the y-axis
V2=∫πx²dy
=∫πe^2ydy
=π/2e^2y|
=π/2(e²-1).

Find the area of plane figure enclosed by curve y = LNX, straight line y = 1, y = 2 and x = 0

Y = LNX, y = 1, y = 2 and x = 0
And LNY = 1
Y = LNX and y = 2 = = > intersection B (E, 2)
y=lnx ==>x=e^y
S=ʃ(1,2) e^y dy=e^y|(1,2)=e²-e
The area of the plane figure enclosed is e? E

Find the area pole of the plane figure enclosed by the curve y = LNX, the straight line x = 1, y = 1 Find the area pole of the plane figure surrounded by curve y = LNX, straight line x = 1, y = 1, and the volume of the body of revolution generated by one revolution around the X axis Thank you very much for your speed Please.

The area of the enclosed plane figure = ∫ (1-lnx) DX = x (1-lnx) │ + ∫ DX (using the method of partial integration) = - 1 + (E-1) = E-2, the volume generated by one revolution around the x-axis = ∫ π (1-LN? X) DX = π [x (1-lnx) │ + 2 ∫ lnxdx] (using the method of partial integration) = π [- 1 + 2 (xlnx │ - ∫ DX)]

The volume of the body of revolution is obtained by rotating the plane figure surrounded by the curve y = LNX and the straight line y = 0 and x = E Don't write me the answer directly

According to the title, we can draw the plane diagram surrounded by the curve y = LNX and the straight line y = 0 and x = e. the oblique side is the right angle area of the curve, and the range of the triangle x is 1 to e, and the range of Y is 0 to 1. Then: the area edge part y = LNX, x = e ^ y (inverse function), since the bottom surface of the rotated object is circular, the volume can be calculated by the annular surface

Calculate the area of the plane figure enclosed by the curve y = x square + 1 and X axis, Y axis and x = 1

The area of the plane figure enclosed by y = x ^ 2 + 1 and X axis, Y axis and x = 1 is = ∫ < 0,1 > (x ^ 2 + 1) DX = [(x ^ 3) / 3 + x] | < 0,1 > = 1 / 3 + 1 = 4 / 3

Find the area s of the plane figure enclosed by the curve y = 1-x square and X axis= How to find the plane area enclosed by two curves?

The intersection point of Y and X is (- 1,0) (1,0)
Then s = ∫ [- 1,1] YDX
=∫[-1,1](1-x^2)dx
=x-x³/3[-1,1]
=4/3

The area s of the plane figure surrounded by the square 2 of the curve y = x and the square 2 of x = y, and the volume V of the body of revolution obtained by the plane figure rotating around the X axis for one cycle are calculated

S=∫(0,1)[x(1/2)]dx-∫(0,1)[x^2]dx
=[2/3(x^(3/2))-1/3(x^3)](0,1)
=2/3-1/3
=1/3
V=π∫(0,1)[x]dx-π∫(0,1)[x^4]dx
=π[1/2(x^2)-1/5(x^5)](0,1)
=3π/10