Question

How would I solve the improper integral ʃ xe¯ᵡ² from 0 to infinity

Answer 1

thats improper Integral

Answer 2

This is an innocent enough looking integral. However, because infinity is not a real number we can’t just integrate as normal and then “plug in” the infinity to get an answer.

Answer 3

I understand what you mean the integral isn't the trouble its the bounds that's messing with me

Answer 4

When you solve this it comes to - e¯ᵡ² / 2 but how would I evaluate

Answer 5

Length of wire = 10 cm.
Length of Side of Square = s
Length of Side of Triangle = t
Area of Square = A(s)
Area of Triangle = A(t)
Perimeter of Square = Ps
Perimeter of Triangle = Pt
Total Area = A(x)


Ps = 4x

Pt = 10 - Ps
Pt = 10 - 4x

t = 1/3 Pt
t = 1/3 (10 - 4x)

A(x) = x²

A(t) = t² (√3) / 4
A(t) = 0.443 t²
A(t) = 0.433 [1/3 (10 - 4x)]²

A(x) = x² + 0.433 [1/3 (10 - 4x)]²
A(x) = x² + 0.433 (10/3 - 4/3 x)²
A(x) = x² + 0.443 [100/9 - (40/9 x) - (40/9 x) + 16x²/9]
A(x) = x² + 0.443 (100/9 - 80/9 x + 16x²/9)
A(x) = x² + 0.443 (11.11 - 8.88x + 1.78x²)
A(x) = x² + 0.443 (1.78x² - 8.88x + 11.11)
A(x) = x² + 0.79x² - 3.9x + 4.9
A(x) = 1.79x² - 3.9x + 4.9

Converting to Standard Form,

A(x) = (1.79x² - 3.9x) + 4.9
A(x) = 1.79(x² - 2.11x) + 4.9
A(x) = 1.79(x² - 2.11x +1.11) + 4.9 - 1.79(1.11)
A(x) = 1.79(x - 1.06)² + 4.9 - 2
A(x) = 1.79(x - 1.06)² + 2.9
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Diagram: http://i533.photobucket.com/albums/ee339 …

Answer 6

try
\[u=-x^2, du=-2xdx\]

Answer 7

\[-\frac{1}{2}e^{-x^2}\]is the anti derivative,
take the limit as x goes to infinity, get 0, replace x by 0 and get \(-\frac{1}{2}\) so your integrals is \(0-(-\frac{1}{2})=\frac{1}{2}\)

Answer 8

First write x^2 as t then our equation reduces to
\[\int\limits_{0}^{\infty}e^{-t}t^{1/2}dt\]

limit is again from 0 to infinity which is obvious
Now this problem is reduced to taking a Laplace Transform of t^(1/2) for the value of s=1
Hence the solution is
\[\sqrt{pi}/2\]

Answer 9

Thre is no way you can solve this problem without having an idea of laplace transform.
laplace transform of the function f(t) is \[\int\limits_{0}^{\infty}e^{-st}f(t)dt\] where s is a parameter. I used this concept above & stating the fact that Laplace transform of t^(1/2) is
\[\sqrt{pi}/(2*s^{3/2})\]
Value of s is 1 here.