The value of the definite integral `int_2^4x(3-x)(4+x)(6-x)(10-x)+sinx)dxe q u a l s` `cos2+cos4` (b) `cos2-cos4` `sin2+sin4` (d) `sin2-sin4`

Options

A. `cos2+cos4`
B. `cos2-cos4`
C. `sin2+sin4`
D. `sin2-sin4`

Answer

B

Explanation

`I=int_(2)^(4)(x(3-x)(4+x)(6-x)(10-x)+sinx)dx`……………….1
`=int_(2)^(4)((6-x)(3-(6-x))(4+(6-x))(6-(6-x))`
`(10-(6-x))+sin(6-x))dx`
`=int_(2)^(4)((6-x)(x-3)(10-x)x(4+x)+sin(6-x))dx`……….2
Adding 1 and 2 we get
`2I=int_(2)^(4)(sinx+sin(6-x))dx`
`=(-cosx+cos(6-x))_(2)^(4)`
`=-cos4+cos2+cos2-cos4`
`=2(cos2-cos4)`
or `I=cos2-cos4`