The area bounded by the curve y = f(x),the x- | vedclass

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(C) The area bounded by the curve $y = f(x)$ from $x = 1$ to $x = b$ is given by the integral $\int_{1}^{b} f(x) \, dx = (b - 1) \sin(3b + 4)$.Replaci

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$\sin(3x + 4) + 3(x - 1) \cos(3x + 4)$

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(C) The area bounded by the curve $y = f(x)$ from $x = 1$ to $x = b$ is given by the integral $\int_{1}^{b} f(x) \, dx = (b - 1) \sin(3b + 4)$.Replacing $b$ with $x$,we get the area function $A(x) = \int_{1}^{x} f(t) \, dt = (x - 1) \sin(3x + 4)$.By the Fundamental Theorem of Calculus,we differentiate both sides with respect to $x$ to find $f(x)$:$f(x) = \frac{d}{dx} [(x - 1) \sin(3x + 4)]$.Using the product rule $\frac{d}{dx} [u \cdot v] = u'v + uv'$:$f(x) = \frac{d}{dx}(x - 1) \cdot \sin(3x + 4) + (x - 1) \cdot \frac{d}{dx} \sin(3x + 4)$.$f(x) = 1 \cdot \sin(3x + 4) + (x - 1) \cdot \cos(3x + 4) \cdot 3$.$f(x) = \sin(3x + 4) + 3(x - 1) \cos(3x + 4)$.

The area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = b$ is $(b - 1) \sin(3b + 4)$. Then $f(x)$ is:

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