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Let the definite integral be defined by the formula `int_(a)^(b)f(x)dx=(b-a)/2(f(a)+f(b))`. For more accurate result, for `c epsilon (a,b), ` we can use `int_(a)^(b)f(x)dx=int_(a)^(c)f(x)dx+int_(c)^(b)f(x)dx=F(c)` so that for `c=(a+b)/2` we get `int_(a)^(b)f(x)dx=(b-a)/4(f(a)+f(b)+2f(c))`.
If `f''(x)lt0 AA x epsilon (a,b)` and `c` is `a` point such that `altcltb`, and `(c,f(c))` is the point lying on the curve for which `F(c)` is maximum then `f'(c)` is equal to

A

`(f(b)-f(a))/(b-a)`

B

`(2(f(b)-f(a)))/(b-a)`

C

`(2f(b)-f(a))/(2b-a)`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
B

`f''(x)lt0AAxepsilon(a,b)`, for `cepsilon(a,b)`
`F(c)=(c-a)/2(f(a))+(f(c))+(b-c)/2(f(b)+f(c))`
`=(b-a)/2f(c)+(c-a)/2f(a)+(b-c)/2f(b)`
or `F'(c)=(b-a)/2f'(c)+1/2f(a)-1/2f(b)`
`=1/2[(b-a)f'(c)+f(a)-f(b)]`
`F''(c)=1/2(b-a)f''(c)lt0`
`[ :' f''(x)lt0Ax epsilon(a,b)` and `bgta`]
Therefore `F(c)` is maximum at the point `(c,f(c))` where
`F'(c)=0` or `f'(c)=2 ((f(b)-f(a))/(b-a))`
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