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If $A(x)=\left|\begin{array}{ccc}1 & 2 & 3 \\ x+1 & 2 x+1 & 3 x+1 \\ x^2+1 & 2 x^2+1 & 3 x^2+1\end{array}\right|$,
then $\int_0^1 A(x) d x$ equals
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then $\int_0^1 A(x) d x$ equals
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$A(x)=\left|\begin{array}{ccc}1 & 2 & 3 \\ x+1 & 2 x+1 & 3 x+1 \\ x^2+1 & 2 x^2+1 & 3 x^2+1\end{array}\right|$
Apply $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_2$, we get
$$
\begin{aligned}
A(x) & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
x+1 & x & x \\
x^2+1 & x^2 & x^2
\end{array}\right| \\
& =0 \quad\left[\because C_2 \text { and } C_3 \text { are identical }\right] \\
\therefore \quad \int_0^1 A(x) d x & =\int_0^1 0 d x=0
\end{aligned}
$$
Apply $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_2$, we get
$$
\begin{aligned}
A(x) & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
x+1 & x & x \\
x^2+1 & x^2 & x^2
\end{array}\right| \\
& =0 \quad\left[\because C_2 \text { and } C_3 \text { are identical }\right] \\
\therefore \quad \int_0^1 A(x) d x & =\int_0^1 0 d x=0
\end{aligned}
$$
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