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If $f(x)=x^{2}, g(x)=2 x, 0 \leq x \leq 2$ then the value of $\mathrm{I}(x)=\int_{0}^{2} \max (f(x), g(x))$ is
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32
\begin{array}{|c|c|c|}
\hlinex & 0 & 2 \\
\hliner(x) & 0 & 16 \\
\hline
\end{array}
Let $r(x)=f(x) \cdot g(x)$
$=x^{2} \cdot 2 x=2 x^{3}$
$r^{\prime}(x)=6 x^{2}$
Put $6 x^{2}=0, \therefore x=0$
$\operatorname{Max} r(x)=2(2)^{3}=16$
or Max $(f(x), g(x))=16$
$I(x)=\int_{0}^{2} 16 d x$
$I(x)=[16 x]_{0}^{2}=32-0=32$
\hlinex & 0 & 2 \\
\hliner(x) & 0 & 16 \\
\hline
\end{array}
Let $r(x)=f(x) \cdot g(x)$
$=x^{2} \cdot 2 x=2 x^{3}$
$r^{\prime}(x)=6 x^{2}$
Put $6 x^{2}=0, \therefore x=0$
$\operatorname{Max} r(x)=2(2)^{3}=16$
or Max $(f(x), g(x))=16$
$I(x)=\int_{0}^{2} 16 d x$
$I(x)=[16 x]_{0}^{2}=32-0=32$
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