The area bounded by the curves `y=f(x),` the x-axis, and the ordinates `x=1a n dx=b` is `(b-1)sin(3b+4)dot` Then `f(x)` is. `(x-1)cos(3x+4)` `sin(3x+4)` `sin(3x+4)+3(x-1)cos(3x+4)` None of these

Explanation

`\int_{1}^{b} f(x) d x=(b-1) \sin (3 b+4)`

we will differentiate both side with respect to b
`f(b) \cdot 1-0=(b-1) \cdot 3 \cos (3 b+4)+\sin (3 b+4) \cdot 1`


`f(b)=3(b-1) \{cos}(3 b+4)+\sin (3 b+4)`


`f(x)=3(x-1) \{Cos}(3 x+4)+\sin (3 x+4)`