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(A) The anti-derivative of $\sin 2x - 4e^{3x}$ is a function $F(x)$ such that $F'(x) = \sin 2x - 4e^{3x}$.We know that $\frac{d}{dx}(\cos 2x) = -2 \si

Vedclass · Accepted Answer

$-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}$

Vedclass · Answer

(A) The anti-derivative of $\sin 2x - 4e^{3x}$ is a function $F(x)$ such that $F'(x) = \sin 2x - 4e^{3x}$.We know that $\frac{d}{dx}(\cos 2x) = -2 \sin 2x$,which implies $\frac{d}{dx}(-\frac{1}{2} \cos 2x) = \sin 2x$.We also know that $\frac{d}{dx}(e^{3x}) = 3e^{3x}$,which implies $\frac{d}{dx}(-\frac{4}{3} e^{3x}) = -4e^{3x}$.Combining these,we have $\frac{d}{dx}(-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}) = \sin 2x - 4e^{3x}$.Therefore,the anti-derivative is $-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}$.

Find an anti-derivative (or integral) of the function $\sin 2x - 4e^{3x}$ by the method of inspection.

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