Answer
$\approx 0.4769$
Work Step by Step
Given: $I=\int_{0}^{1} x(1.1)^{-x^2} \ dx$
Let us consider that $u=-x^2 \implies dx=-\dfrac{du}{2x}$
Now, we have $I= \int_{0}^{1} x(1.1)^{u} (-\dfrac{du}{2x})$
or, $=-\dfrac{1}{2} \int_{0}^{1} (1.1)^{u} \ du$
or, $=[\dfrac{-(1.1)^{-x^2}}{2 \ln 1.1}]_{0}^{1} +C$
or, $=[\dfrac{-(1.1)^{-(1)^2}}{2 \ln 1.1}]+[\dfrac{-(1.1)^{-(0)^2}}{2 \ln 1.1}]$
or, $=\dfrac{1-1.1^{-1}}{2 \ln 1.1}$
or, $\approx 0.4769$
