Given ∫ [0, + ∞] x ^ (- 1 / 2) e ^ (- x) DX = √π, find I = ∫ [- ∞, ∞] x ^ 2E ^ (- x ^ 2) DX Steps to take

Given ∫ [0, + ∞] x ^ (- 1 / 2) e ^ (- x) DX = √π, find I = ∫ [- ∞, ∞] x ^ 2E ^ (- x ^ 2) DX Steps to take

Let x = T ^ 2 (T > 0) in a given condition
Then ∫ (0 → + ∞) e ^ (- T ^ 2) / T * 2tdt = √π
∫(0→+∞)e^(-t^2)dt=√π/2
Because e ^ (- T ^ 2) is an even function
So ∫ (∞→ + ∞) e ^ (- T ^ 2) DT = ∫ (∞→ 0) e ^ (- T ^ 2) DT + ∫ (0 → + ∞) e ^ (- T ^ 2) DT = 2 ∫ (0 → + ∞) e ^ (- T ^ 2) DT = √π
The original formula = - 1 / 2 * ∫ (∞→ + ∞) Xe ^ (- x ^ 2) d (- x ^ 2)
=-1/2*∫(-∞→+∞)xd(e^(-x^2))
=-xe^(-x^2)/2|(-∞→+∞)+1/2*∫(-∞→+∞)e^(-x^2)dx
=0+√π/2
=√π/2

∫ (0 → + ∞) (x ^ 2) (e ^ (- x) DX = 2? Probability
Is it a distribution function value?
￥（3）=2
What do you think

This problem should be to find e (x ^ 2) of exponential distribution when λ = 1

Find the integral C squared (1-x squared) DX (integral range 0 to 1

The original formula = ∫ (0,1) C & sup2; (1-x & sup2;) DX
=C²(x-x³/3)│(0,1)
=C²(1-1/3)
=2C²/3.

Calculate integral ∫ 0 → θ x ^ 2 / θ (1-x / θ) ^ (n-1) DX

∫0→θ x^2/θ(1-x/θ)^(n-1)dx
=θ^2*∫0→1( x^2(1-x)^(n-1)dx)
=θ^2*∫0→1(-1/n*x^2*d((1-x)^n))
=θ^2/n*[-x^2(1-x)^n|(0->1)+∫0→1(2x(1-x)^ndx)]
=2θ^2/(n(n+1))*∫0→1(-xd((1-x)^(n+1)))
=2θ^2/(n(n+1))*[-x(1-x)^(n+1)|(0->1)+∫0→1((1-x)^(n+1)dx)
=2θ^2/(n(n+1)(n+2))

e^x/(1-e^x)dx

∫e^x/(1-e^x)dx
=-∫1/(1-e^x) d(1-e^x)
=-ln|1-e^x| +C

Calculate ∫ upper limit 1 lower limit-1 (x ^ 2 + 2x-3) DX

∫ (from - 1 to 1) (x ^ 2 + 2x-3) DX = [(x ^ 3 / 3) + x ^ 2-3x] (from - 1 to 1) = 1 / 3 + 1-3 - [(- 1 / 3) + 1 + 3] = 1 / 3 + 1-3 + 1 / 3-1-3 = 2 / 3-6 = - 16 / 3

The continuity of function f (x) = ln (e ^ n + x ^ n) / N, (x > 0) is discussed in higher numbers (Tongji version)
This is divided into (0, e] and (E, ∞) discussions. How do you know how to use this method to divide partitions?

Observe the function, when n is determined, where ln part of the function may be discontinuous is less than 0 and the denominator is 0. Obviously, it seems that the problem does not exist
The original formula = ln (e ^ n (1 + (x / E) ^ n)) / N = 1 + ln (1 + (x / E) ^ n) / N, X - > e, N - > ∞
There may be discontinuities

Function problems determined by parameter equation in Higher Mathematics of Tongji Sixth Edition
If x = φ (T) y = ψ (T) determines a function, then y = f (x), where x determines more than y, then Yang becomes a multi valued function. What the book says is a function. How do you understand this?
For example, x = acost, y = asint, which determines an ellipse, is a multi valued function, while the textbook is treated as a single valued function

I have studied this problem. Generally speaking, many valued functions can always be regarded as several single valued functions, and implicit functions can deal with this kind of problems. Therefore, in general, it is not necessary to distinguish between many valued functions and single valued functions. Moreover, the processing methods in textbooks can be explained by the implicit function rule, which requires us to learn the existence theorem of implicit functions in Volume II

In the chapter of multivariate function of Tongji sixth edition of higher mathematics, the definition of closed region is the point set of open region and boundary?
For example, if a non open and non closed set is connected, it can not be called an open region by definition. Right? Since it is not called an open region, its boundary is not called a closed region by this definition. The conclusion is obviously wrong. I think the definition of a closed region should be changed to a connected closed set

A non open and non closed set is certainly not a region, but its closure (that is, union upper boundary) is not necessarily a region
It's a closed area. It may or may not be
The definition means that a set that can be expressed as a closure of an open region is a closed region
So your conclusion is obviously wrong. I don't know where to find it?
A connected closed set is not necessarily a closed domain,
For example, {(x, y): y = SiNx, 0

Monotone bounded sequence must have limit
Monotonic boundedness includes upper bound and lower bound. How can there be a limit? What is the limit

Boundedness includes upper and lower bounds. When increasing, the limit is upper bound, and when decreasing, the limit is lower bound