Chapter 7: Integrals

Ans:

$\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right) $  $=\int{{{x}^{\frac{1}{2}}}dx}+\int{{{x}^{-\frac{1}{2}}}}dx=\frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}}+\frac{{{x}^{\frac{1}{2}}}}{\frac{1}{2}}+C $ 

$=\frac{2}{3}{{x}^{\frac{3}{2}}}+2{{x}^{\frac{1}{2}}}+C $ 

Thus, the correct answer is C.

Ans: Given, $\frac{d}{dx}f\left( x \right)=4{{x}^{3}}-\frac{3}{{{x}^{4}}}$

Anti-derivative of $4{{x}^{3}}-\frac{3}{{{x}^{4}}}=f\left( x \right)$

$\therefore f\left( x \right)=\int{4{{x}^{3}}-\frac{3}{{{x}^{4}}}=f\left( x \right)} $ 

$f\left( x \right)=4\int{{{x}^{3}}dx-3\int{\left( {{x}^{-4}} \right)}dx} $ 

$f\left( x \right)=4\left( \frac{{{x}^{4}}}{4} \right)-3\left( \frac{{{x}^{-3}}}{-3} \right)+C $ 

$f\left( x \right)={{x}^{4}}+\frac{1}{{{x}^{3}}}+C $ 

$Also, $ 

$f\left( 2 \right)=0 $ 

$\therefore f\left( 2 \right)={{\left( 2 \right)}^{4}}+\frac{1}{{{\left( 2 \right)}^{3}}}+C=0 $ 

$\Rightarrow 16+\frac{1}{8}+C=0 $ 

$\Rightarrow C=\frac{-129}{8} $ 

$\therefore f\left( x \right)={{x}^{4}}+\frac{1}{{{x}^{3}}}-\frac{129}{8} $ 

Thus, the correct answer is A.