Question

Find the following integrals.
If $\frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}$ such that f(2)=0. Then f(x)is
$\left(a\right)x^4+\frac{1}{x^3}-\frac{129}{8}\left(b\right)x^3+\frac{1}{x^4}+\frac{129}{8}\left(c\right)x^4+\frac{1}{x^3}+\frac{129}{8}\left(d\right)x^3+\frac{1}{x^4}-\frac{129}{8}$

Answer 1

Given, $\frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}$
$\Rightarrow Anti-derivativeof(4x^3-\frac{3}{x^4})=f\left(x\right)\therefore f\left(x\right)=\int (4x^3-\frac{3}{x^4})dx=4\int x^{3}dx-3\int x^{-4}dx\Rightarrow f\left(x\right)4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C=x^4+\frac{1}{x^3}+C$
Also, $f\left(2\right)=0,\therefore f\left(2\right)=(2{)}^4+\frac{1}{(2{)}^3}+C$
$\Rightarrow 16+\frac{1}{8}+C=0\Rightarrow C=-(16+\frac{1}{8})\Rightarrow C=-\frac{129}{8}$
$\therefore f\left(x\right)=x^4+\frac{1}{x^3}-\frac{129}{8}$. Hence, the correct option is (a).