Question

I need help solving this definite integral, from x = -2 to x=e: f(x)=x-e, if x>0 and f(x)=e^(x+2), if x<=0
Thanks !

Answer 1

you need to split this up into two integrals
where are you having trouble?

Answer 2

I got final answer being -2+e^(e+3). I tried to check it but it was wrong according to geogebra :S

Answer 3

is that first function\(f(x)=x-e\) or \(f(x)=x^{-e}\) ?

Answer 4

\[f(X)=x-e, if x>0 \\ f(x)=e^{x+2}, if x\le 0\]

Answer 5

so what did you put for the first integral?

Answer 6

we split into integral fropm -2 to near 0 (x-e) and from 0 to e (e^(x+2))

Answer 7

good, so\[\int_{-2}^0x-edx+\int_0^ee^{x+2}dx\]right?

Answer 8

exactly

Answer 9

\[\int_{-2}^0x-edx+\int_0^ee^{x+2}dx=\frac{x^2}2-ex|_{-2}^0+e^{x+2}|_0^e=2+2e+e^{e+2}-e^2\]

Answer 10

I tried to check, it is not checking, doing the calculation in geogebra, or wolfram alpha