The area enclosed by the curve y root number x, the tangent y x 2 and the y axis is

y=root number x, and the intersection point with y=x-2 is (4,2).

Calculated with dx: a= (below 0 on 4) ( x-x+2) dx = [(2 3) x -x 2+2x] (below 0 on 4) = 16 3

When using dy, note that the inverse function of y=root x is x=y (y 0), and the integral can only be found above the x-axis of the graph, that is, from 0 to 2, and the triangle below the x-axis must be added.

a= (lower 0 upper 2) (y+2-y )dy+1 2 2 2=[y 2+2y-y 3](lower 0 upper 2) + 2=10 3+2=16 3

There is also a simple method: the area below the x-axis of the graph is equal to the area of the graph enclosed by y=x-2, x=4 and x-axis, so the area of the graph is the area of the graph enclosed by y=root number x, x=4 and x-axis, thus.

a= (lower 0 upper 4) xdx = [(2 3) x ] (lower 0 upper 4) = 16 3

First, the intersection point of the curve and the straight line is calculated, and the simultaneous equations x=y 2 and y=x-2 are solved to obtain the intersection points (4,2) (rounded) and (1,-1).

Cut into two pieces with y=-1, top.

Then the integral formula of the curve is x(integral)=(1 3)y3, then the area from the origin to the point (1,-1) is s1=1, below 3. The area from the intersection point to the intersection point of the line and the y-axis is s2=(1 2)*1*1=1 2, then the total area is s=s1+s2=5 6

The intersection of the curve y= x and the straight line y=x-2 is (4,2), so the area of the graph = (0,4)[ x-x+2]dx

2 3)x (3 Hu 2)-(1 2)x 2+2x| (0,4)

Calculate the area of the graph enclosed by the line y=x-4, the curve y= (2x), and the x-axis.

Solution 1: Let (x-4) = 2x), get x -8x+16=2x, that is, there is x -10x+16=(x-2)(x-8)=0, get x =2, x =8;

Since y= (2x) 0, the curve has no graph below the x-axis, so x=2 should be rounded; There is only one intersection point between a straight line and a curve (8,4).

Then let y=x-4=0 to get the intersection of the line and the x-axis (4,0);

Therefore, according to the area of the figure enclosed by the title, s=[0,8] 2x) dx-(1 2) (8-4) 4=[(2 2) 3]x (3 2) [0,8]- 8

Solution 2: You can also integrate y, in this case you should write y=(2x) as x=y 2;Write y=x-4 as x=y+4;Rule.

The area of the enclosed figure s=[0,4] (y+4-y 2)dy=[y 2+4y-y 6][0,4]=8+16-64 6=24-32 3=40 3,3,

First, solve the intersection point a (-4(1+2),-4(2+2)) of two straight lines y=2*x and y=x-4

If we find out the intersection point b(4,0) of the line y=x-4 and the x-axis, then AOB is sought in the Qing Dynasty.

s=1/2*4*|-4(2+√2)|=8(2+√2)

Summary. The area is 1

Known curve y=xFind the area of the tangent of the curve at x=1 and the area enclosed by the x-axis.

The area is 1 good.

Emergency. Is that so?

The tangent and the x-axis do not form a closed area, and I think I need to add an x 1<> I don't know, so I asked.

The question is not to ask the area of the besieged city with the coordinate axis <>

It's just that two lines can't be counted.

The area is infinity.

Vocabulary questions. The first.

Ah, I see, and curves.

Solve the system of equations and draw a sketch. Then determine the best line for the Jacky Joke score.

Given two situations, the difference is the same way.

For reference, please laugh Navu do.