Answer
$$
\lim _{t \rightarrow 0} \left[ \int_{0}^{1}[b x+a(1-x)]^{t} d x \right]^{\frac{1}{t}} =e^{-1}(\frac{b^{b}}{a^{a}})^{\frac{1}{(b-a)}}
$$
Work Step by Step
$$
\begin{aligned}
\int_{0}^{1}[b x+a(1-x)]^{t} d x& =\int_{a}^{b} \frac{u^{t}}{(b-a)} d u \\
&\quad\quad\quad\left[\begin{array}{c}{ \text {Let }u=b x+a(1-x), x : 0 \rightarrow 1 } \\ { \text {Then }d u=(b-a) d x}, u:a \rightarrow b \end{array}\right] \\
&=\left[\frac{u^{t+1}}{(t+1)(b-a)}\right]_{a}^{b} \\
&=\frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}.
\end{aligned}
$$
Let
$$
y=\lim _{t \rightarrow 0}\left[ \frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}\right]^{\frac{1}{t}}
$$
Then
$$
\ln y= \lim _{t \rightarrow 0} \left[ {\frac{1}{t}} \ln \frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}\right]
$$
The limit is of the form $\frac {0}{0} $, so we can use L'Hospital's rule to get:
$$
\begin{aligned}
\ln y &=\lim _{t \rightarrow 0}\left[\frac{b^{t+1} \ln b-a^{t+1} \ln a}{b^{t+1}-a^{t+1}}-\frac{1}{t+1}\right] \\
&=\frac{b \ln b-a \ln a}{b-a}-1 \\
&=\frac{b \ln b}{b-a}-\frac{a \ln a}{b-a}-\ln e \\
&=\ln \frac{b^{b /(b-a)}}{e a^{a /(b-a)}}
\end{aligned}
$$
Thus
$$
y=e^{-1}(\frac{b^{b}}{a^{a}})^{\frac{1}{(b-a)}},
$$
and therefore
$$
\lim _{t \rightarrow 0} \left[ \int_{0}^{1}[b x+a(1-x)]^{t} d x \right]^{\frac{1}{t}} =e^{-1}(\frac{b^{b}}{a^{a}})^{\frac{1}{(b-a)}}
$$
