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

Question

Dr. Abubacker · Accepted Answer

(c)
$\because\ \int_{1}^{b}{f(x)dx = (b - 1)\sin{(3b + 4)}}$
$	herefore$ On differentiating both sides with respect to $b$, we get
$f(b) = 3(b - 1)\cos{(3b + 4) + \sin{(3b + 4)}}$
$	herefore\ f(x) = 3(x - 1)\cos{(3x + 4) + \sin{(3x + 4)}}$

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

Explanation

Reviewed by Dr. Abubacker

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