If frac{d[f(x)]}{dx} = g(x) for a le x le b,t | vedclass

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(C) Let $I = \int_a^b f(x)g(x) dx$. Given that $\frac{d[f(x)]}{dx} = g(x)$,we can write $g(x) dx = d[f(x)]$. Substituting $f(x) = t$,we get $df(x) = d

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$\frac{[f(b)]^2 - [f(a)]^2}{2}$

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(C) Let $I = \int_a^b f(x)g(x) dx$. Given that $\frac{d[f(x)]}{dx} = g(x)$,we can write $g(x) dx = d[f(x)]$. Substituting $f(x) = t$,we get $df(x) = dt$,which implies $g(x) dx = dt$. When $x = a$,$t = f(a)$,and when $x = b$,$t = f(b)$. Thus,the integral becomes $I = \int_{f(a)}^{f(b)} t dt$. Evaluating the integral,we get $I = \left[ \frac{t^2}{2} \right]_{f(a)}^{f(b)} = \frac{[f(b)]^2 - [f(a)]^2}{2}$.

If $\frac{d[f(x)]}{dx} = g(x)$ for $a \le x \le b$,then $\int_a^b f(x)g(x) dx$ equals

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