If the solve the problem


If $f^{\prime}(x)=8 x^{3}-2 x, f(2)=8$, find $f(x)$


Given $f^{\prime}(x)=8 x^{3}-2 x$ and $f(2)=8$

On integrating the given equation, we have

$\int f^{\prime}(x) d x=\int\left(8 x^{3}-2 x\right) d x$

We know $\int \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{f}(\mathrm{x})$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int\left(8 \mathrm{x}^{3}-2 \mathrm{x}\right) \mathrm{dx}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int 8 \mathrm{x}^{3} \mathrm{dx}-\int 2 \mathrm{xdx}$

$\Rightarrow \mathrm{f}(\mathrm{x})=8 \int \mathrm{x}^{3} \mathrm{dx}-2 \int \mathrm{xdx}$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow \mathrm{f}(\mathrm{x})=8\left(\frac{\mathrm{x}^{3+1}}{3+1}\right)-2\left(\frac{\mathrm{x}^{1+1}}{1+1}\right)+\mathrm{c}$

$\Rightarrow \mathrm{f}(\mathrm{x})=8\left(\frac{\mathrm{x}^{4}}{4}\right)-2\left(\frac{\mathrm{x}^{2}}{2}\right)+\mathrm{c}$

$\Rightarrow f(x)=2 x^{4}-x^{2}+c$

On substituting $x=2$ in $f(x)$, we get


$\Rightarrow 8=32-4+c$

$\Rightarrow 8=28+c$

$\therefore c=-20$

On substituting the value of $c$ in $f(x)$, we get

$f(x)=2 x^{4}-x^{2}+(-20)$

$\therefore f(x)=2 x^{4}-x^{2}-20$

Thus, $f(x)=2 x^{4}-x^{2}-20$

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