Question

integral from -1 to 1 of e^(u+1) du.

Answer 1

\[e^{u+1}=e\times e^u\] is a start

Answer 2

use substitution rule
s = u +1
now take the derivative of u + 1
ds = du

e^(s)ds

Answer 3

\[e\int_{-1}^1e^udu\] anti derivative is \(e^u\) evaluate at 1 and -1 and subtract

Answer 4

then use the fundamental theory F(b) - F(a)
where in your case b = 1, and a = -1

Answer 5

how do you know that e^(u+1)=e * e^u

Answer 6

wait nevermind

Answer 7

laws of exponents
\[b^n\times b^m=b^{n+m}\]
\[e^1\times e^u=e^{u+1}\]

Answer 8

can't you jsut do a substitution?

Answer 9

you get as a final answer \(e(e^1-e^{-1})=e^2-1\)

Answer 10

sure you can, but why bother?

Answer 11

I am just so confused with this problem, answer in back of book is 0

Answer 12

its less complicated

Answer 13

not zero for sure

Answer 14

i guess it is a matter of taste
you can use a substitution or you can use \(\int cf(x)dx=c\int f(x)dx\) either way

Answer 15

ok so

use substitution rule
s = u +1
now take the derivative of u + 1
ds = du

thus we have

\[\int\limits_{-1}^{1} e^{s}ds\]
the antiderivative of e^(x) = e^(x)

so we end up with


e^(u+1) as the derivative
using the rule F(b) - F(a) we can find the area
where a = -1 and b = 1
so

e^(1 + 1) - e^(-1 + 1)
=

e^(2) - 1

I think your book is incorrect
https://www.wolframalpha.com/input/?i=ntegral+from+-1+to+1+e^%28u%2B1%29

Answer 16

okay thanks the substitution makes sense now