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If $\frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{f}(\mathrm{x})\}=\mathrm{g}(\mathrm{x})$, then $\int_a^b \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2}\left[\mathrm{f}^2(\mathrm{~b})-\mathrm{f}^2(\mathrm{a})\right]$
Hints $: f(x)=\int g(x) d x$
$$
\begin{aligned}
& \int_a^b f(x) \cdot g(x) \cdot d x=(f(x) f(x))_a^b-\int_a^b g(x) f(x) d x \\
& I=f^2(b)-f^n(a)^{-1} \\
& I=\frac{1}{2}\left(f^2(b)-f^2(a)\right)
\end{aligned}
$$
$$
\begin{aligned}
& \int_a^b f(x) \cdot g(x) \cdot d x=(f(x) f(x))_a^b-\int_a^b g(x) f(x) d x \\
& I=f^2(b)-f^n(a)^{-1} \\
& I=\frac{1}{2}\left(f^2(b)-f^2(a)\right)
\end{aligned}
$$
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