how to integrate e^(1/3*x)?

Question

integral from 2 to 6 of e^(1/3*x)dx

Answer 1

Use substitution letting u = (1/3)x
You will see that the integral of 3^((1/3)x) dx is
3e^((1/3)x)
Plug in 6 for x
3e^((1/3)6) = 3e^2
Plug in 2 for x
3e^((1/3)2) = 3e^(2/3)
Subtract
Answer: 3e^2 - 3e^(2/3)

Note: In general, the integral of e^(kx) dx is (1/k)e^(kx) + C, where x does not equal 0.
You can prove this using substitution.

Answer 2

this is a case of the general form of the function e^x, whose integral is e^x

we can reduce the given function to this form by substituting the exponent to y, ie. letting y=1/3 x

hence dy = 1/3 dx or dx = 3dy

hence integral (e^(1/3*x) dx = integral (e^y 3 dy)
= 3 integral (e^y dy) = 3e^y + constant

now put back the value for y to get

integration = 3e^(1/3*x) + c

now apply the limits 2 to 6 (here c wont matter):

value = 3 (e^2 - e^(2/3))

Answer 3

integral e^(1/3x)=3e^(1/3x)
applying the limits
=3[e^2-e^(2/3)]