English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

evaluate the integral from 0 to 1 at x^2(sqrt[4]{e^x}) dx

To clarify the notation of the problem you are solving. The number 4 after the sqrt means its ^4sqrt(e^x) . It is the fourth root of e^x

2007-11-16 13:03:09 · 4 answers · asked by sportzblock 1 in Science & Mathematics Mathematics

4 answers

to make it better to follow i am change the equation to:

integration[ x^2*e^(x/4) dx] from 0 to 1

udv = uv - integration(vdu)
[to remember: ultraviolet voodoo]

first choose u and dv
u = x^2
dv = e^(x/4) dx
du = 2x*dx
v = 4*e^(x/4)

integration[ x^2*e^(x/4) ] = x^2*4*e^(x/4) - integration [4*e^(x/4)*2x dx ]

4*x^2*e^(x/4) - integration [8x*e^(x/4) dx ]

since there is another integration then we need another by parts for the second integration.

u = 8x
dv = e^(x/4) dx
du = 8 dx
v = 4*e^(x/4)

so now the whole equation is:

4*x^2*e^(x/4) - (8x*4*e^(x/4) - integration[ 4*e^(x/4)* 8 dx]

4*x^2*e^(x/4) - 32x*e^(x/4) - integration[ 32*e^(x/4) dx]

then integrate one last time:

4*x^2*e^(x/4) - (8x*4*e^(x/4) - 32*4*e^(x/4)

4*x^2*e^(x/4) - (8x*4*e^(x/4) - 128*e^(x/4)

now you can plug in 0 and 1

(4*1^2*e^(1/4) - (8*1*4*e^(1/4) - 128*e^(1/4)) - (4*0^2*e^(0/4) - (8*0*4*e^(0/4) - 128*e^(0/4))

= 100* e^(1/4) - 128 = 0.40254166877

2007-11-16 13:19:07 · answer #1 · answered by Anonymous · 0 0

I'll do the indefinite integral, then plug in the limits.
You have to integrate
∫ x² e^(x/4) dx.
So, let u = x/4
x = 4u x² = 16u² dx = 4du
We get
64 ∫ u² e^u du.
We could use integration by parts here, but it's simpler
to use Tic-tac-toe:
u² e^u
2u e^u
2 e^u
0 e^u
So our final result is
64e^u(u² - 2u + 2) = 64e^(x/4)(x²/16 -x/2 +2).
Now plug in the limits:
x = 1: 64 e^(1/4)(25/16)
x = 0: 64(2) = 128
So our final result is 100e^(1/4) - 128.

2007-11-16 21:28:32 · answer #2 · answered by steiner1745 7 · 0 0

write the function as (x^2)e^(x/4)
u=x^2 and v'=e^(x/4) so v=4e^(x/4)
you do integration by parts two times with the same choice for v and you get 4(x^2)e^(x/4)-32xe^(x/4)=128e(x/4) from 0 to 1
It is equal 100e^(1/4)-128

2007-11-16 21:20:41 · answer #3 · answered by Lumi 2 · 0 0

∫x^2e^x dx
= x^2e^x - ∫2xe^x, integration by parts
= x^2e^x - 2xe^x + 2e^x

∫x^2e^(x/4) dx
= 64∫u^2e^u du, where u = x/4
= 64[(x/4)^2e^(x/4) - 2xe^(x/4) + 2e^(x/4)], x from 0 to 1
= 4e^(1/4) - 128

2007-11-16 21:18:41 · answer #4 · answered by sahsjing 7 · 0 0

fedest.com, questions and answers