Question

The value of the definite integral $\underset{2}{\overset{4}{\int }}(x(3-x)(4+x)(6-x)(10-x)+\sin x)dx$ equals

• A. $\cos 2+\cos 4$
  
• B. $\cos 2-\cos 4$
  
• C. $\cos 4-\cos 2$
  
• D. $\sin 2-\sin 4$

Answer 1

The correct option is B $\cos 2-\cos 4$


We have
$I=\underset{2}{\overset{4}{\int }}(x(3-x)(4+x)(6-x)(10-x)+\sin x)dx\cdots \left(1\right)$
Now, replace $x$ with $6-x$
$I=\underset{2}{\overset{4}{\int }}(6-x\left)\right(3-(6-x)\left)\right(4+(6-x)\left)\right(6-(6-x)\left)\right(10-(6-x))+\sin (6-x))dx$

$=\underset{2}{\overset{4}{\int }}((6-x)(x-3)(10-x)x(4+x)+\sin (6-x))dx\cdots \left(2\right)$

On adding $\left(1\right)$ and $\left(2\right),$ we get
$2I=\underset{2}{\overset{4}{\int }}(\sin x+\sin (6-x))dx$
$={[-\cos x+\cos (6-x\left)\right]}_2^4$
$=-\cos 4+\cos 2+\cos 2-\cos 4$
$=2(\cos 2-\cos 4)$
Hence, $I=\cos 2-\cos 4$