- #1

jayanthd

- 16

- 0

**I know ∫(xe[itex]^{ax}[/itex]) dx = x (e[itex]^{ax}[/itex] / a) - (1/a) ∫e[itex]^{ax}[/itex] . 1 dx**

= x (e[itex]^{ax}[/itex] / a) - (1/a) (e[itex]^{ax}[/itex] / a)

= (e[itex]^{ax}[/itex] / a) (x - 1/a)

= x (e[itex]^{ax}[/itex] / a) - (1/a) (e[itex]^{ax}[/itex] / a)

= (e[itex]^{ax}[/itex] / a) (x - 1/a)

i.e, integral of two functions = (first function) (integral of second function) - ∫(integral of second function) (differential of first function)

This is not a homework. I am a working professional and I need help in solving a problem.

The solution I need is for

(1/0.1) ∫20te[itex]^{-10t}[/itex] dt between limits 0 and 20 us. limits can be taken as 0 to t. I don't need numerical solution.

t = [itex]\tau[/itex]

dt = d[itex]\tau[/itex]

integral becomes

200 ∫[itex]\tau[/itex]e[itex]^{-10\tau}[/itex] d[itex]\tau[/itex] between limits 0 and t

it becomes 200 [ [itex]\tau[/itex] (e[itex]^{-10\tau}[/itex] / - 10) + (1/10) ∫e[itex]^{-10\tau} [/itex] . 1 d[itex]\tau[/itex]

= 200 [ [itex]\tau[/itex] (e[itex]^{-10\tau}[/itex] / - 10) + (1/10) (e[itex]^{-10\tau}[/itex] / - 10)]

= 200 [ [itex]\tau[/itex] (e[itex]^{-10\tau}[/itex] / - 10) - (1/100) e[itex]^{-10\tau}[/itex]]

I know I have to apply limits to the two e[itex]^{-10\tau}[/itex]

I want to know should I apply limits also to [itex]\tau[/itex] which is at the beginning of the solution (here... = 200 [ [itex]\tau[/itex] ...) ?

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