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How to find the integral of the fourth power of Sint multiplied by the square of cost?
=(1/4)(1/2)(1/2) ∫(1-cos4t)(1-cos2t)dt
=(1/16) ∫(1-cos4t-cos2t+cos4tcos2t)dt
=t/16-(sin4t)/64-(sin2t)/32+(1/16)(1/2)∫(cos6t+cos2t)dt
=t/16-(sin4t)/64-(sin2t)/32+(1/192)*(sin6t)+(sin2t)/64+C
The cubic power of x = cost and the cubic power of y = sint
No, it's the cubic power of x = acost, the cubic power of y = asint to find their area
x=acost3 y=asint3 x*y=acost3*asint3=a2*(sint*cost)3=a2*(1/2sin2t)3=1/8a2*(sin2t)3
Integration of higher number problems ∫ (t-sint) ^ 2sintdt
The solution is as follows: ∫ (t-sint) ^ 2sintdt = ∫ (T ^ 2sint + Sint ^ 2sint-2tsint ^ 2) DT = ∫ T ^ 2sintdt + ∫ (1-cost ^ 2) sintdt-2 ∫ TSINT ^ 2DT = - ∫ T ^ 2dcost - ∫ (1-cost ^ 2) dcost - ∫ t * (1-cos2t) DT = - T ^ 2cost + ∫ 2cost + cost ^ 3 / 3-T ^ 2 / 2 + ∫ TDS
How to integrate Sint / DT
Integral without DX as denominator
High number: how to calculate the integral of Sint ^ 2cost ^ 7?
S is the integral sign, and brackets are the integral limit
S[0,Pi/2](sint^2*cost^7)dt=?
The answer is (2-1)! (7-1)! / (9-1)!. I really want to know how to get the integral answer,
(2-1)! (7-1)
RT.sin2t (Sint + cost) DT solution
I can't type the symbol in front of me.
The symbol in front can't be typed
The definition of function f (x) on the symmetry of image point (a, 0)
Such as the title
The function f (x) is symmetric with respect to the image point (a, 0),
Then any point on f (x) image is symmetric with respect to (a, 0)
It is still on the f (x) image,
Where f (x) satisfies f (2a-x) + F (x) = 0
Which of the following is right?
A: F (x) represents the algebraic expression B containing X: the range of function is the book B in its definition
C: Function is a special mapping D: mapping is a special function
Why is wrong wrong? What should be changed
C
If the domain of F (x + 1) is [- 2,3], then the domain of y = f (2x-1) is? I don't need an answer. I want to tell you the difference between the domain of definition and the range of value from this question. What's the relationship between F (x + 1) and f (2x-1) and f (x)
The definition field is the value range of the independent variable x, and the value range is the value range of the function value y
The domain of definition and the corresponding law f determine the range of value, that is, as long as the domain of definition and the corresponding law f are determined, the range of value will be determined!
Correspondence rule can be understood as the rule of operation. For the same correspondence rule, the requirements for the following quantities are the same
For example, the rule of correspondence is to take the reciprocal, which requires that these numbers are not zero
F (x) = 1 / x, then x ≠ 0, f (x + 1) = 1 / (x + 1), then x + 1 ≠ 0
For example, the corresponding rule is open root, which requires these numbers to be greater than or equal to 0
F (x) = √ x, then x ≥ 0, f (2-3x) = √ (2-3x) then 2-3x ≥ 0
Your topic:
In this kind of problem, we should pay special attention to that the domain of definition must be the scope of X
The definition field of y = f (x + 1) is [- 2,3], which refers to the range of X in it, that is - 2 ≤ x ≤ 3, from which - 1 ≤ x + 1 ≤ 4 can be obtained
The domain of definition of y = f (2x-1) is also the scope of X in it. Because the corresponding rules are all F, they are the same
So the range of quantity after f is the same, that is, the range of X + 1 and 2x-1 are the same
So the solution of - 1 ≤ 2x-1 ≤ 4 is 0 ≤ x ≤ 5 / 2
So the domain of definition is [0,5 / 2]
F (x + 1) and f (2x-1) and f (x)
The corresponding rules of the three are the same, all F
So the ranges of X + 1,2x-1, X in the three brackets are the same
However, it should be noted that the domains of these three functions refer to the scope of X, not necessarily the overall scope in brackets
1: It is known that the domain of F (x) is the domain of [3,15] to find f (X & # 178; - 2x)
2: If the domain of function f (X & # 178;) is [- 1,1], find the domain of F (x + 1)
Please compare the similarities and differences between these two questions. My main question is: 1. The independent variables of these functions are not clear. 2. The answer to the first question is [- 3, - 1] ∪ [3,5], but why do we need to solve the inequality system of X & # 178; - 2x ≥ 3 in the process?
x²-2x≤15
First of all, let's be clear: domain refers to the scope of X
The key to this kind of problem is to grasp f (..), the range of the formula in brackets is invariable
It is generally divided into the following steps:
1) From the definition field given by the title, the scope of the formula in brackets can be obtained
2) From the law that the range of the formula in brackets remains unchanged, we can get the range of another formula in brackets
3) Solving inequality
As you said in question 1:
1) From the definition field of F (x) as [3,15], we get that the range of the formula in "bracket" (in this case, X itself) is [3,15]
2) The range of x ^ 2-2x is [3,15]
3) Solving inequality
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