Ask/Help sape terer add math masuk sini

[mohdfauzi5599]

Wat L I A T E
settle kan cos then Exponent.
Try find calculus book.

 

[ezxury]

alamak sumpah x paham...adesh...
saya x tahu add math langsung nie, tdi buat2 pandai je ngn adik..hahha

soalan: integrate following function with respext to x

soalan 1:http://latex.codecogs.com/gif.latex?e^2^x%20cos%20x
soalan 2:http://latex.codecogs.com/gif.latex?x^2%20e^4^x

 

[irsaffy]

Nah refer sini..))

http://www.youtube.com/watch?v=8uFQeJiXC5M

 

[tokputih]

The idea is to integrate by parts twice. Start with the original integral,
∫sin(x)e^(2x) dx =

Let u = e^(2x) and dv = sin(x). Then du = 2e^(2x) and v = -cos(x). Apply integration by parts to get:

∫u dv = uv - ∫v du
∫e^(2x)sin(x) dx = -e^(2x)cos(x) - ∫-cos(x)2e^(2x) dx
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx

Call this last equation *1*. Now this seems to have made the integral we couldn't compute, ∫e^(2x)sin(x) dx, into yet another integral we can't compute, namely ∫cos(x)e^(2x) dx. So, apply integration by parts again - this time on just that rightmost integral.

For ∫cos(x)e^(2x) dx, let u = e^(2x) and dv = cos(x). Then du = 2e^(2x) and v = sin(x). So, from integration by parts you get:

∫u dv = uv - ∫v du
∫e^(2x)cos(x) dx = sin(x)e^(2x) - ∫sin(x)2e^(2x) dx
∫e^(2x)cos(x) dx = sin(x)e^(2x) - 2∫sin(x)e^(2x) dx

Now plug this substitution of the integral ∫e^(2x)cos(x) dx into equation *1* and see if it becomes cleaner.
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2[sin(x)e^(2x) - 2∫sin(x)e^(2x) dx]
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x) - 4∫sin(x)e^(2x) dx
∫e^(2x)sin(x) dx + 4∫sin(x)e^(2x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)
5∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)
5∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))
∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))/5

Hope this helps you!

 

[ezxury]

huhu sangat2 berharap dpt tahu jalan kerje die..ekekke semangat nk blaja pown ade

 

[wan ridzuan]

Ni la salah satu sebab aku suka tidoq waktu add math dulu I-) I-).....serabut giler.... ) )

 

[TT]

jawapannya C. 3.02y

 

[VeilSupra]

fuhh den suke soalan DE ni zaman2 spm dulu

 

[ezxury]

The idea is to integrate by parts twice. Start with the original integral,
∫sin(x)e^(2x) dx =

Let u = e^(2x) and dv = sin(x). Then du = 2e^(2x) and v = -cos(x). Apply integration by parts to get:

∫u dv = uv - ∫v du
∫e^(2x)sin(x) dx = -e^(2x)cos(x) - ∫-cos(x)2e^(2x) dx
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx

Call this last equation *1*. Now this seems to have made the integral we couldn't compute, ∫e^(2x)sin(x) dx, into yet another integral we can't compute, namely ∫cos(x)e^(2x) dx. So, apply integration by parts again - this time on just that rightmost integral.

For ∫cos(x)e^(2x) dx, let u = e^(2x) and dv = cos(x). Then du = 2e^(2x) and v = sin(x). So, from integration by parts you get:

∫u dv = uv - ∫v du
∫e^(2x)cos(x) dx = sin(x)e^(2x) - ∫sin(x)2e^(2x) dx
∫e^(2x)cos(x) dx = sin(x)e^(2x) - 2∫sin(x)e^(2x) dx

Now plug this substitution of the integral ∫e^(2x)cos(x) dx into equation *1* and see if it becomes cleaner.
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2[sin(x)e^(2x) - 2∫sin(x)e^(2x) dx]
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x) - 4∫sin(x)e^(2x) dx
∫e^(2x)sin(x) dx + 4∫sin(x)e^(2x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)
5∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)
5∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))
∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))/5

Hope this helps you!

sedikit membantu thx

 

[qubs]

senang je nie pakai kalkulator je