·What is the area of the closed plane figure enclosed by the curve y = x ^ 2, the straight line y = 2x-1 and the X axis?

·What is the area of the closed plane figure enclosed by the curve y = x ^ 2, the straight line y = 2x-1 and the X axis?

Calculus problems, high school science students will

Find the area of the plane figure enclosed by the square of the curve y = x and the straight line y = 2x + 5

The coordinates of intersection point of parabola and straight line are (1 - √ 6,7-2 √ 6), (1 + √ 6,7 + 2 √ 6),
Enclosed area s = ∫ (1 - √ 6 → 1 + √ 6) (2x + 5) DX - ∫ (1 - √ 6 → 1 + √ 6) x ^ 2DX
=(x^2+5x-x^3/3)（1-√6→1+√6）
=8√6.

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

from
y＝x2
The coordinates of the intersection point y = x are (0,0), (1,1),
from
y＝x2
We get (y, 2, x, x, x, x, x, x, x, x, y, x, x, y, x, x, y, x, y, x, y, x, y, x, x, y, x, y, x, y, X, y, x, y, x, y, x, y, X (2 points)
∫ the area s is s = ∫
One
Zero
(2x−x)dx+∫
Two
One
(2x−x2)dx… (6 points)
=∫
One
Zero
xdx+∫
Two
One
(2x−x2)dx=x2
2|
One
Zero
+(x2−x3
3)
Two
One
=7
6… (10 points)

Find the area of the plane figure enclosed by the sum of squares of the following curve y = 3-x and the straight line y = 2x

The first step is to find the intersection point, (0,0), (1,1)
The second step is to integrate later
∫ (x-xx) DX, integral interval is (0,1)
Obviously ∫ (x-xx) DX = 0.5xx - (XXX / 3), substituting the value, we get
Area = 1 / 6

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

Because it's a symmetric graph, only the area where x > 0 is required
At this time, the graph is surrounded by curves x = √ y, x = √ (Y / 2), y = 1
Then, for the y-integral, the area of this part = ∫ [0,1] (√ Y - √ (Y / 2)) dy
=∫[0,1](1-√2/2)√ydy
=(1-√2/2)(2/3)
=(2-√2)/3
The area of the plane figure obtained is 2 (2 - √ 2) / 3

The area of plane figure enclosed by curve: y = LNX, Y axis and straight line y = LNA, y = LNB (b > a > 0)

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The volume of the plane figure surrounded by the curves y = LNX, x = E and y = 0 is Step by step

(E-2)

Find out the area s of the plane figure enclosed by the curve y = x ^ 2-2x, y = 0, x = 1, x = 3, and calculate the volume V of the rotating body obtained by the plane figure revolving around the X axis

S=2,V=pi*46/15
The detailed process is shown in the figure below

The volume of the sphere is generated by the curve y = LNX, x = e, y = 0

It's a ring
The upper limit is 1 and the lower limit is 0
The curves surrounding the graph are y = LNX, x = e ^ y and x = E
Volume v = π∫ (0 to 1) [(E) 2 - (e ^ y) 2] dy
=π∫ (0 to 1) [e? - e ^ (2Y)] dy
=π * [e? Y - (1 / 2) e ^ (2Y)], (0 to 1)
= π*[e² - (1/2)e² + (1/2)]
= (π/2)(1+e²)

The volume of (x, x) is the area of the curve formed by the rotation of (x, x) and (x) around the axis of (1) Detailed steps are required. Thank you

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