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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 find $f(x).$

What should i do in this question,I tried taking $f(x)=A\sin(3x+4)+B\cos(3x+4)$ and put this in equation $\int_{1}^{b}f(x)dx=(b-1)\sin(3b+4)$ and tried to find $A$ and $B$.But this method failed.How should i find $f(x)?$

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  • $\begingroup$ It is simply not possible to find the $f$, because there are infinitely many functions $f$ satisfying the assumptions. But to find one $f$, you can simply differentiate $(b-1)\sin(3b+4)$ wrt $b$. Another (easier) $f$ would be a constant function $f(x)=a$ with an appropriate constant $a$... $\endgroup$
    – sranthrop
    Commented Sep 27, 2015 at 3:19

2 Answers 2

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Hint: Remember that $\int_{a}^{b}f'(x)dx = f(b) - f(a).$

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HINT: area bounded by the curve $y=f(x)$, the x-axis, between $x=1$ & $x=b$ is $$A=\int_{1}^{b}f(x)\ dx$$ But the area bounded is $A=(b-1)\sin(3b+4)$, hence we get $$\int_{1}^{b}f(x)\ dx=(b-1)\sin(3b+4)$$ Substituting $b=x$, $$\int_{1}^{x}f(x)\ dx=(x-1)\sin(3x+4)$$

Now, Use F.T.C. to find $f(x)$

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