Let `a+b=4,w h e r ea<2,a n dl e tg(x)` be a differentiable function. If `(dg)/(dx)>0` for all `x ,` prove that `int_0^ag(x)dx+int_0^bg(x)dxin c r e r a s e sa s(b-a)in c r e r a s e sdot`

Explanation

`a+b=4` or `b=4-a`
Let `f'(a)=int_(0)^(a)g(x)dx+int_(0)^(b)g(x)dx`
`=int_(0)^(a)g(x)dx+int_(0)^(4-a)g(x)dx`
`:.(df(a))/(da)=g(a)-g(4-a)`
Now, `(df(a))/(d(b-a))=(df(a))/(d(4-2a))=(df(a))/(-2da)=(g(4-a)-g(a))//2`
given `alt2`
or `2alt4`
or `4-agta`
or `g(4-a)gt g(a) [ :' (dg(x))/(dx) gt 0`. So `g(x)` is an increasing function]
Thus, `(df(a))/(d(b-a))gt0`
Therefore, `f(a)=int_(0)^(a)g(x)dx+int_(0)^(a)g(x)dx` increases as `(b-a)` increases.