Not complex at all

$\begin{aligned} \int_0^1 (-3)^x dx & = \int_0^1 (3\color{#3D99F6}{e^{i \pi}})^x dx \quad \quad \small \color{#3D99F6}{e^{i \pi} = \cos \pi + i \sin \pi = -1} \\ & = \int_0^1 (\color{#3D99F6}{e^{\ln{3}}}e^{i \pi})^x dx \quad \quad \small \color{#3D99F6}{e^{\ln 3}=3} \\ & = \int_0^1 e^{(\ln{3} +i \pi)x} dx \\ & = \left[ \frac{e^{(\ln{3} +i \pi)x}}{\ln{3} +i \pi} \right]_0^1 = \left[ \frac{(-3)^x}{\ln{3} +i \pi} \right]_0^1 = \frac{(-3)^1 - (-3)^0}{\ln{3} +i \pi} = \frac{-4}{\ln{3} +i \pi} \\ & = \frac{-4 \ln{3}+ i4\pi}{(\ln{3})^2+\pi^2} = - 0.396734346 + i 1.13450188 \end{aligned}$

$\Rightarrow \lfloor 100ab \rfloor = \lfloor 45.00958615 \rfloor = \boxed{45}$

First note than $\int { a^{x}dx }=\frac{a^x}{\ln{a}}$ then $\int_{0} ^{1} { (-3)^{x}dx }=\frac{(-3)^x}{\ln{(-3)}}$ This evaluates to $\frac{-4}{\ln{-3}}=-\frac{4}{\ln{3}\ln{-1}}$ , then from euler's identity i have that -\frac { 4 }{ \ln { 3 } +i\pi } =-\frac { 4 }{ \ln { 3 } +i \pi } \frac { \ln { 3-i \pi } }{ \ln { 3 } -i \pi } =\frac { -4\ln { 3 } +4i \pi }{ \ln { ^{ 2 } } { 3 }+\pi ^{ 2 } } =-\frac {4\ln { 3 } }{ \ln { ^{ 2 } } { 3 }+\pi ^{ 2 } } +\frac { 4\pi }{ \ln { ^{ 2 } } { 3 }+\pi ^{ 2 } } i now a= $\frac { 4\ln{3} }{ \ln { ^{ 2 } } { 3 }+\pi ^{ 2 } }$ , b= $\frac { 4\pi }{ \ln { ^{ 2 } } { 3 }+\pi ^{ 2 } }$ and $\left\lfloor 100ab \right\rfloor =45$

-3=3(CosPI+SinPI) Apply De Moivre's theorem

Let say $u = (-3)^x$ and we can have $du = (-3)^x (\ln(3)+i \pi) dx$ or $\frac{du}{ \ln(3)+i \pi}= (-3)^x dx :$

So,

$\begin{array}{lcl}\int_{0}^{1}(-3)^{x} dx &=&\int_{1}^{-3} \frac{du}{ \ln(3)+i \pi}\\ &= &\frac{u}{\ln(3)+i \pi}|_{1}^{-3} \\ &= &\frac{-4}{\ln(3)+i \pi} \\ &=& \frac{(-4)(\ln(3)-i\pi)}{(\ln(3))^2-(i \pi)^2} \\&=& \frac{(-4)(\ln(3))}{(\ln(3))^2+\pi^2} + \frac{4\pi}{(\ln(3))^2+\pi^2}i \end{array}$

$\implies \big\lfloor 100 * {\frac{(4)(\ln(3))}{(\ln(3))^2+\pi^2} * \frac{4\pi}{(\ln(3))^2+\pi^2}}\big\rfloor=\boxed{45}$