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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 !
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you need to split this up into two integrals where are you having trouble?
I got final answer being -2+e^(e+3). I tried to check it but it was wrong according to geogebra :S
is that first function\(f(x)=x-e\) or \(f(x)=x^{-e}\) ?
\[f(X)=x-e, if x>0 \\ f(x)=e^{x+2}, if x\le 0\]
so what did you put for the first integral?
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we split into integral fropm -2 to near 0 (x-e) and from 0 to e (e^(x+2))
good, so\[\int_{-2}^0x-edx+\int_0^ee^{x+2}dx\]right?
exactly
\[\int_{-2}^0x-edx+\int_0^ee^{x+2}dx=\frac{x^2}2-ex|_{-2}^0+e^{x+2}|_0^e=2+2e+e^{e+2}-e^2\]
I tried to check, it is not checking, doing the calculation in geogebra, or wolfram alpha
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