How to calculate the area of a graph by calculus? The area of the figure enclosed in the first quadrant by the X, Y axis and the circle with (a, b) as the center and radius R Arithmetic square root with radius less than (A2 + B2)~

It is found that the condition of LZ is too broad. It is equal to finding the area of the circle and the first quadrant in any case
It should be solved according to the situation

Find the graphic area enclosed by the image of function y = x and function y = x ^ 2 (square). (the unit is a unit length on the coordinate axis)

Integral: first find 1 / 2 of the triangle area, and then subtract the integral of X'2 on 0 to 1

How to use calculus to find the area between X square + y square = 4 and straight line y = 1

First define the integral symbol: s (a, b) f (x) DX is the definite integral of F (x) in the interval of x = (a, b). The lower limit is a and the upper limit is B
x^2 + y^2 = 4
The interval of F (x) is calculated by x ^ 2 + y ^ 2 = 4 and y = 1, a = - √ 3, B = √ 3
Here is the definite integral:
Judging from the value range of X, y is greater than 0, so: y = √ (4 - x ^ 2) = 2 √ [1 - (x / 2) ^ 2]
S = S(a,b)f(x)dx
It is a little troublesome to do the integral operation, and the original function can be transformed into a parametric equation
x = R*cost
y = R*sint
R = 2
A = - √ 3, corresponding to T1 = π / 6
B = + √ 3, corresponding to T2 = π - π / 6 = 5 π / 6
The integral is transformed into polar coordinates
The sector of the differential area element is similar to a triangle. The height of the triangle is the circle radius R and the bottom edge is RDT
dS = (R*Rdt)/2
S = (1/2)*S(t1,t2)(R^2)*dt
= (R^2/2)*t |(t1,t2)
= (R^2/2)*(t2 - t1)
= (R^2/2)*(5π/6 - π/6)
= (R^2/2)*(2π/3)
= (πR^2)/3 .R = 2
= 4π/3

As shown in the figure, it is the part of the image with quadratic function y = - 1 / 2x ^ 2 above the x-axis. What is the area enclosed by this image and the x-axis? Please use calculus And, make it clear. Some don't understand. And say, what's the formula?

You can't write that expression clearly. I'll just give you the idea:
1. Calculate the integral of the function
2. The interval of F (x) > 0 is calculated
3. The area is obtained by substituting the end value of the interval into the integral

The area of the closed graph surrounded by the image of function y = x-x2 and the X axis is equal to __

By equations
y＝0
y＝x−x2 ， The solution shows that X1 = 0, X2 = 1
Therefore, the area of the figure is s = ∫
one
0
（ x-x2）dx
=（1
2x2-1
3x3）|
one
0
=1
6．
So the answer is: 1
6．

3、 Fill in the blanks (5 sub questions in total, 4 points for each sub question), 11. 11. The definition field of the function is 12． . 13. Limit 14. Set, then _________ 15. The derivative of the function is 4、 Calculation questions (4 sub questions in total, 40 points in total) 16. Find the limit 17. 16. Request 18. Find the derivative of the function determined by the implicit function equation 19. Find the extreme value of the function

Don't copy it directly from word. The title can't be displayed

How to calculate the area of a graph by calculus? The area of the figure enclosed in the first quadrant by the X, Y axis and the circle with (a, b) as the center and radius R Arithmetic square root with radius less than (A2 + B2)~