Help integrating e^rt x ? Thanks!?

Question

Any help would be appreciated!

e ^ sqrt (x) evaluated between 1 and 4

Answer 1

let u² = x then you have 2u e^u du...integration by parts

Answer 2

∫ e^(√x) dx =

let √x = t →

x = t²→

dx = 2t dt

thus, substituting, you get:

∫ e^(√x) dx = ∫ e^t  2t dt =

2 ∫ t e^t dt =

this is a typical by parts integral:

let t = u → dt = du

e^t dt = dv → e^t = v

thus, integrating by parts, you get:

2 ∫ t e^t dt = 2 [t e^t - ∫ e^t dt] =

2t e^t - 2 ∫ e^t dt =

2t e^t - 2 e^t + C =

that is, susbstituting back t = √x,

2√x e^(√x) - 2 e^(√x) + C  (the antiderivative)

finally, evaluate the definite integral from 1 to 4, that is:

4

∫ e^(√x) dx = [2√4 e^(√4) - 2 e^(√4)] - [2√1 e^(√1) - 2 e^(√1)] =

1

[2(2) e^(2) - 2 e^(2)] - (2e - 2e) =

4e^(2) - 2e^(2) - 0 = 2e²

I hope it helps...

Bye!

Answer 3

e^sqrt(x) would be equal to e^(x^1/2)

let u=x^1/2  let dv=e^sqrt x

du=1/2x^-1/2     v=e^sqrt x

=uv- integral vdu

=x^1/2 (e sqrt x)- 1/2 integral e sqrt x (x^-1/2)

then let w=sqrt x

dw= 1/2 sqrt x^-1/2

making the equation

=x^1/2 (e sqrt x)- e sqrt x + C