The curvilinear trapezoid represents the figure limited to the schedule of non-negative and continuous function f on an interval [a; b], axis of OX and straight lines of x=a and x=b. For calculation of its area use a formula: (S=Fb) – F(a), where F – an antiderivative for f.
It is required to you
- - pencil;
- - handle;
- - ruler.
Instruction
1. You need to define area of the curvilinear trapezoid limited to a function graph of f (x). Find primitive F for the set function f. Construct a curvilinear trapezoid.
2. Find several control points for function f, calculate coordinates of crossing of the schedule of this function with OX axis if they are available. Represent graphically other set lines. Shade a required figure. Find x=a and x=b. Calculate the area of a curvilinear trapezoid, using a formula (S=Fb) – F(a).
3. Example of I. Determine the area of a curvilinear trapezoid, the limited line y=3x-x². Find an antiderivative for function y=3x-x². It will be F(x) =3/2x²-1/3x³. Function y=3x-x² represents a parabola. Its branches are directed down. Find points of intersection of this curve with OX axis.
4. From the equation: 3x-x²=0, follows that x=0 and x=3. Required points – (0; 0) and (0; 3). Therefore, a=0, b=3. Find some more control points and represent the schedule of this function. Calculate the area of the set figure on a formula: (S=Fb) – F(a) = F(3)-F(0)=27/2-27/3-0+0=13.5-9=4.5.
5. Example of II. Determine the area of the figure limited to lines: y=x² and y=4x. Find antiderivatives for these functions. It will be F(x) =1/3x³ for the y=x function² and G(x) =2x² for the y=4x function. By means of the system of the equations find coordinates of points of crossings of a parabola of y=x² and the linear y=4x function. Such points two: (0;0) and (4;16).
6. Find control points and represent schedules of the set functions. It is easy to notice that the required area is equal to the difference of two figures: the triangle formed by direct y=4x, y=0, x=0 and x=16 and the curvilinear trapezoid limited to the y=x lines², y=0, x=0 and x=16.
7. Calculate the areas of these figures on a formula: S¹=G(b) – G(a) = G(4)-G(0)=32-0=32 and S²=F(b) – F(a) = F(4)–F(0)=64/3–0=64/3. So, the area of a required figure of S is equal to S ¹-S² =32–64/3=32/3.