Question

How the heck do I go about solving:
integral 2te^((t^2)/6)

Answer 1

try letting u = t^2 or 1/6 t^2

Answer 2

if we do u= t^2
then du = 2 t dt
and \[ 2 t e^{\frac{t^2}{6}} dt = e^{\frac{u }{6}} du \]

Answer 3

if you still are unsure, let v = u/6
dv = 1/ 6 du
and
\[ e^{\frac{u}{6} } du = 6 e^{\frac{u}{6} } \frac{1}{6} du \\
6 e^v dv
\]

Answer 4

Ya I got integral e^u/6 which becomes 6e^(t^2/6)+c. But it's a definite integral from 4 to 10 so the answer should be: 6e^(100/6)-6e^(16/6) I believe...thanks for the help!

Answer 5

Yep phi is correct. nice job phi.