Giải ∫ (from 0 to 1) of xe^-x^2 wrt x | Ứng dụng giải toán Microsoft Math

Tính giá trị

- \frac{1}{2 e} + \frac{1}{2} \approx 0.316060279

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Bài kiểm tra

5 bài toán tương tự với:

\int_{0}^{1} x e^{- x^{2}} d x

Các bài toán tương tự từ Tìm kiếm web

How to evaluate

\int_{0}^{1} e^{- x^{2}} d x

using power series?

https://math.stackexchange.com/questions/2404105/how-to-evaluate-int-01-mathrm-e-x2-mathrm-dx-using-power-series

Just collecting the material in the comments and converting it into answer. We have

e^{- x^{2}} = \sum_{n \geq 0} \frac{( - 1 ) ^{n} x ^{2 n}}{n !}

where the series in the RHS is absolutely convergent for any

x \in R

\int_{0}^{x} e^{- x^{2}} d x

https://www.quora.com/What-is-displaystyle-int_0-x-e-x-2-dx

What is the upper limit of integration ? I assume the upper limit as infinity. To solve then, substitute x^2=t So, the integral gets simplified to Gamma function form and it is evaluated as ...

\int_{0}^{1} (1 + e^{- x^{2}}) d x

https://math.stackexchange.com/questions/1087369/integral-int-011e-x2-dx

\int_{0}^{1} 1 + e^{- x^{2}} = \int_{0}^{1} 1 + \int_{0}^{1} e^{- x^{2}} = 1 + \int_{0}^{1} e^{- x^{2}}

\int e^{- x^{2}} = \frac{π erf(x)}{2} ⟹ I = 1 + \frac{π erf ( 1 )}{2} \approx 1.746

erf(x) = \frac{2}{π} \int_{0}^{x} e^{- t^{2}} d t

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

https://math.stackexchange.com/questions/494958/use-the-method-of-cylindrical-shells-to-find-the-volume-generated-by-rotating-th

It looks as if you may have reached the right integral, which is

\int_{0}^{1} 2 π x e^{- x^{2}} d x .

You can quickly integrate by making the substitution

u = x^{2}

\int_{0}^{1} e^{- x^{2}} d x

https://math.stackexchange.com/questions/65681/how-to-compute-int-01-e-x2-dx

e r f (z)

is the "error function" encountered in integrating the normal distribution (which is a normalized form of the Gaussian function). It is an entire function defined by

e r f (z) = \frac{2}{π} \int_{0}^{z} e^{- t^{2}} d t

How we compare the two following integral without calculation?

https://math.stackexchange.com/questions/847612/how-we-compare-the-two-following-integral-without-calculation/847622

x^{2} \leq x

x \in [0, 1]

e^{x^{2}} \leq e^{x}

x \leq x

x (x - 1) \leq 0

x \in [0, 1]

x e^{x^{2}} \leq x e^{x}

\int_{0}^{1} x e^{x^{2}} d x \leq \int_{0}^{1} x e^{x} d x

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Phương trình bậc hai

x^{2} - 4 x - 5 = 0

4 sin θ cos θ = 2 sin θ

Phương trình tuyến tính

y = 3 x + 4

699 * 533

[2534] [2 - 1 0135]

Phương trình đồng thời

{8 x + 2 y = 46 7 x + 3 y = 47

\frac{d}{d x} \frac{( 3 x ^{2} - 2 )}{( x - 5 )}

\int_{0}^{1} x e^{- x^{2}} d x

x \to - 3 lim \frac{x ^{2} - 9}{x ^{2} + 2 x - 3}