If an antiderivative of f(x) is e^x and that | vedclass

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(C) Given that the antiderivative of $f(x)$ is $e^x$,we have $f(x) = \frac{d}{dx}(e^x) = e^x$.Given that the antiderivative of $g(x)$ is $\cos x$,we h

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$e^x \cos x + c$

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(C) Given that the antiderivative of $f(x)$ is $e^x$,we have $f(x) = \frac{d}{dx}(e^x) = e^x$.Given that the antiderivative of $g(x)$ is $\cos x$,we have $g(x) = \frac{d}{dx}(\cos x) = -\sin x$.We need to evaluate the integral $I = \int f(x) \cos x \, dx + \int g(x) e^x \, dx$.Substituting the values of $f(x)$ and $g(x)$:$I = \int e^x \cos x \, dx + \int (-\sin x) e^x \, dx$.$I = \int e^x (\cos x - \sin x) \, dx$.Recall the standard integral formula $\int e^x (h(x) + h'(x)) \, dx = e^x h(x) + c$.Here,let $h(x) = \cos x$,then $h'(x) = -\sin x$.Thus,$I = e^x \cos x + c$.

If an antiderivative of $f(x)$ is $e^x$ and that of $g(x)$ is $\cos x,$ then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx = $

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