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

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(A) The area bounded by the curve $y = f(x)$,the $x-$ axis,and the ordinates $x = 1$ and $x = b$ is given by the integral $\int_1^b f(x) \, dx$.Accord

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The area bounded by the curve $y = f(x)$,the $x-$ axis,and the ordinates $x = 1$ and $x = b$ is given by the integral $\int_1^b f(x) \, dx$.According to the problem,the area is given by $\int_1^b f(x) \, dx = (b - 1)\sin(3b + 4)$.To find $f(x)$,we differentiate both sides of the equation with respect to $b$ using the Leibniz Rule (Fundamental Theorem of Calculus):$\frac{d}{db} \left( \int_1^b f(x) \, dx \right) = \frac{d}{db} [(b - 1)\sin(3b + 4)]$.Applying the product rule on the right side:$f(b) = \frac{d}{db}(b - 1) \cdot \sin(3b + 4) + (b - 1) \cdot \frac{d}{db}(\sin(3b + 4))$.$f(b) = 1 \cdot \sin(3b + 4) + (b - 1) \cdot \cos(3b + 4) \cdot 3$.$f(b) = 3(b - 1)\cos(3b + 4) + \sin(3b + 4)$.Replacing $b$ with $x$,we get $f(x) = 3(x - 1)\cos(3x + 4) + \sin(3x + 4)$.

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

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