Answer
$3-\displaystyle \frac{3}{e}$
Work Step by Step
FTC: Let $f$ be a continuous function defined on the interval $[a, b]$,
and let $F$ be any antiderivative of f defined on $[a, b]$. Then
$\displaystyle \int_{a}^{b}f(x)dx=\left[F(x)\right]_{a}^{b}=F(b)-F(a)$.
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$f(x)=3e^{x}$, which is continuous on [$-1,0$].
First, find an antiderivative $F(x)$:
$\displaystyle \int f(x)dx=3\int e^{x}dx=3e^{x}+C$
(we take C=0, since by FTC we need ANY antiderivative)
$F(x)=3e^{x}$
Next, evaluate:
$F(0)=3$
$F(-1)=\displaystyle \frac{3}{e}$
Finally, apply the theorem:
$\displaystyle \int_{-1}^{0}f(x)dx=F(0)-F(-1)=3-\frac{3}{e}$
