Question

If the area bounded by the curve $y=f\left(x\right)$ and the $x-$axis from $x=\frac{1}{3}$ to $x=b$, where $b>0$ is $(3b^2-2b+\frac{1}{3})$.sq.unit, and the function be $f\left(x\right)=6x+m$.

Then what will be the value of $-m$?

Answer 1

$\because {\int }_{\frac{1}{3}}^{b}f\left(x\right)dx=3b^2-2b+\frac{1}{3}$ .......... $\left(1\right)$ (given)

Differentiating both sides w.r.t $b$, we get

$f\left(b\right)-f\left(\frac{1}{3}\right)=3b^2-2b+\frac{1}{3}$........(applying leibnitz rule)

$\Rightarrow f\left(b\right)=6b-2$

$\Rightarrow f\left(x\right)=6x-2$

Note: $f\left(\frac{1}{3}\right)=3{\left(\frac{1}{3}\right)}^2-2\left(\frac{1}{3}\right)+\frac{1}{3}=\frac{1}{3}-\frac{2}{3}+\frac{1}{3}=0$

Hence, the value of $m$ is $-2$.