find the area under the curve y=(above the x-axis) from x=0 to x=2
given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region
let g(x)=cos x^2 , f(x)=x^1/2 and a and b (a<b) be the root of the quadratic equation 18x^2-9(pi)x+(pi)^2=0 then the area bounded by the curve y=(gof)(x)and the lines x=a , x=b and y=0 is. let a (x, y) be the point on the curve y = f(x). then slope of the tangent to the curve at point a is `"dy"/"dx"`. according to the given condition, `"dy"/"dx" = "x" + 3"y" - 1` ∴ `"dy"/"dx" - 3"y" = "x - 1"` ....(1) this is the linear differential equation of the form `"dy"/"dx" + "py" = "q"`, where p = - 3 and q = x - 1 ∴ i.f. = `"e"^(int "p dx") = "e"^(int - 3"dx") = "e"^(- 3"x")` ∴ the solution of (1) is given by `"y" * ("i.f.") = int "q" * ("i.f.") "dx" + "c"` ∴ `"y" * "e"^(- 3"x") = int ("x - 1") * "e"^(-3"x") "dx" + "c"` ∴ `"e"^(- 3"x") * "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") * int "e"^(- 3"x") "dx"] "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = ("x - 1") * "e"^(- 3"x")/-3 - int 1 * "e"^(- 3"x") * "y"/-3 "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") * "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 * "e"^(- 3"x")/-3 + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"\_1` ∴ 9y = - 3(x - 1) - 1 + 9`"c"\_1 * e^("3x")` ∴ 9y + 3(x - 1) + 1 = `9"c"\_1 * e^("3x")` ∴ 9y + 3x - 3 + 1 = `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `"c" * e^("3x")` where c = 9c1 ....(2) this is the general equation of the curve. but the required curve is passing through the origin (0, 0). ∴ by putting x = 0 and y= 0 in (2), we get 0 = 2 + c ∴ c = - 2 ∴ from (2), the equation of the required curve is 3(x + 3y) = 2 - 2`e^("3x")` i.e. 3(x + 3y) = 2(1 - `e^("3x")`). 圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872668990.jpg
 
首先請先 複習下列公式
 
設扇形的中心角θ弳，半徑為r，弧長為s，面積為a，則
s=rθ；a= r2θ= rs
sol: 設: ∠aob=θ 弳 (弧度) od=x
(x+2.5) θ =6 ……………………………..(1)
求黃色部分面積= (x+5)2 θ - x2θ
= θ ( x2+10x+25-x2 )
= θ (10x+25)
=5θ( x + 2.5)=30



圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669011.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669012.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669033.jpg
 



 2010-06-04 23:25:06 補充：
y= |cosx|

恕不再多畫 將以文字說明

1. 當cosx >0 時 y=cosx
2. 當cosx <0 時 y = -cosx

 2010-06-04 23:26:21 補充：
注意

y=3sin2(x-π)+2

之 a , b 步驟為同一圖形

 2010-06-05 13:06:49 補充：
奇怪 1/2 通通不見了

更正

s=rθ；a=0.5 r2θ=0.5 rs

求黃色部分面積=0.5(x+5)2 θ - 0.5x2 θ
=0.5θ ( x2+10x+25-x2 )
=θ (10x+25)
=5θ( x + 2.5)=30

p.s. x2 即為 x平方

 2010-08-28 11:35:07 補充：
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rid of several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2 1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 } the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points. May as well promote my comment to an answer.


You need to add points to a `Curve2D` object, then adjust the control points of the curve, and then finally draw the curve. So here's some simple, almost pseudo code:



```
var array_of_line_points # This already has the vectors which describe our line
for point in array_of_line_points:
  # The "get_perpendicular_vector()" function returns a vector that's a copy of the point, yet has been slid along a line parallel to two neighboring points. The "distance" variable is how far the control point should be from the originating point.
  var control_point1 = get_perpendicular_vector(point, distance)
  var control_point2 = get_perpendicular_vector(point, -distance)
  curve.add_point(point, control_point1, control_point2)
# ...In the draw function
func _draw():
  draw_polyline(curve.get_baked_points(), red, 2.0)

```

I got the inspiration for this from [Rob Spencer's page on Spline Interpolation](http://scaledinnovation.com/analytics/splines/aboutSplines.html). I'd like to make this more thorough (i.e. defining the `get_perpendicular_vector()` function). But this is the best I can currently do. ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17.