Question

Did i do this right?

Answer 1

\[\int\limits_{-1}^{1} (e ^{u+1})du = e^{2}-1\]

Answer 2

\[\int\limits_{-1}^1(e^{u+1})\text du=\frac{e^{u+1}}{u+1}\Big|_{-1}^1\]

Answer 3

or did you mean

\[\int\limits_{-1}^1(e^{u}+1)\text du \]

Answer 4

no its e^{u+1}

Answer 5

e^2/2-1/0?

Answer 6

thats what i got, but 1/0 dosent really make sense

Answer 7

maybe we have to take a limit

Answer 8

e^0=1?

Answer 9

d/du e^{u+1} = e^{u+1} * d/du (u+1)

Answer 10

\[e^0 =1 \checkmark\]

Answer 11

so is it not.... e^2/2-1/0?

Answer 12

the zero in the denominator is freaking me out a bit

Answer 13

hahah yeah...

Answer 14

\[\frac{e^{u+1}}{u+1}\Big|_{-1}^1\]\[=\frac{e^{1+1}}{1+1}-\lim\limits_{x\rightarrow-1}\frac{e^{x+1}}{x+1}\]\[=\frac{e^{2}}{2}-\lim\limits_{x\rightarrow-1}\frac{e^{x+1}}{x+1}\]\[\stackrel{\text{l'H}}=\frac{e^{2}}{2}-\lim\limits_{x\rightarrow-1}\frac{(x+1)e^{x+1}}{1}\]

Answer 15

\[\int\limits_{-1}^{1}e^{u+1} du = e^{u+1} |_{-1}^{1}\]

Answer 16

hmm? did you use a subsitution or something @scarydoor ?

Answer 17

\[\frac{d}{du} e^{u+1} = e^{u+1} \frac{d}{du} (u+1) = e^{u+1} * 1 = e^{u+1}\]

Answer 18

so e^(u+1) is an anti-derivative of e^(u+1) (forgetting about the constant...).

So there's no need for the 1/(u+1). I didn't spot that at first either.

Answer 19

What tipped me off, is that if you plot e^(u+1) it'll look something like:

And the area under the curve there really shouldn't involve any kind of limits.