$f\left(x\right)$  and $h\left(x\right)$  are two real valued continuous function and let  $\int h\left(x\right)dx=f^{-1}\left(x\right)$ and $f\left(x\right)=x^3+x+\sin 2\pi x+2$  and the value of $\underset{2}{\overset{4}{\int }}xh\left(x\right)dx$  is $\text{'}k\text{'}$  , then the value of $\text{'}4k\text{'}$  is,

Let  $k=\underset{2}{\overset{4}{\int }}xh\left(x\right)dx$
Using by parts,  $k={\left(x{f}^{-1}\left(x\right)\right)}_2^4-\underset{2}{\overset{4}{\int }}{f}^{-1}\left(x\right)dx=4f^{-1}\left(4\right)-2f^{-1}\left(2\right)-\underset{2}{\overset{4}{\int }}{f}^{-1}\left(x\right)dx$
 $\because f^{-1}\left(4\right)=1,f^{-1}\left(2\right)=0$
We know,  $\underset{a}{\overset{b}{\int }}f\left(x\right)dx+\underset{f\left(a\right)}{\overset{f\left(b\right)}{\int }}{f}^{-1}\left(x\right)dx=bf\left(b\right)-af\left(a\right)$
 $\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{2}{\overset{4}{\int }}{f}^{-1}\left(x\right)dx=1.4-0.2=4$
Now, $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}\left(x^3+x+\sin 2\pi x+2\right)dx$  
 $={\left(\frac{x^4}{4}+\frac{x^2}{2}-\frac{\cos 2\pi x}{2\pi }+2x\right)}_0^1=\left(\frac{1}{4}+\frac{1}{2}-\frac{1}{2\pi }+2\right)-\left(-\frac{1}{2\pi }\right)=\frac{11}{4}$
So,  $\frac{11}{4}+\underset{2}{\overset{4}{\int }}{f}^{-1}\left(x\right)dx=4\Rightarrow \underset{2}{\overset{4}{\int }}{f}^{-1}\left(x\right)dx=\frac{5}{4}$
Hence,  $k=4-\frac{5}{4}=\frac{11}{4}$