Question

Let f(x) be a non-negative continuous function such that the area bounded by the curve y = f(x), x-axis and the ordinates x = (π/4) and x = β > π/4 is(β sin β + (π/4) cos β + √2 β). Then f (π/2) is :

Answer 1

area of closed curve = integrate { f(x) } limit [π/4 , ß]

ßsinß +(π/4)cosß+√2ß) =[g(x) +C ] limit [π/4 , ß]

g(ß) -g(π/4) =ßsinß +(π/4)cosß +√2ß

it means
g(x ) -g(π/4) =xsinx +π/4cosx +√2x

put x =0
g(π/4 )= -π/4
hence ,
g(x ) =xsinx +π/4 cosx +√2x -π/4

now ,
x =π/2 put
g(π/2 ) = π/2 +0 +π/√2 -π/4
=π/4 +π/√2

Answer 2

Answer:

Step-by-step explanation:

area of closed curve = integrate { f(x) } limit [π/4 , ß]

ßsinß +(π/4)cosß+√2ß) =[g(x) +C ] limit [π/4 , ß]

g(ß) -g(π/4) =ßsinß +(π/4)cosß +√2ß

it means

g(x ) -g(π/4) =xsinx +π/4cosx +√2x

put x =0

g(π/4 )= -π/4

hence ,

g(x ) =xsinx +π/4 cosx +√2x -π/4

now ,

x =π/2 put

g(π/2 ) = π/2 +0 +π/√2 -π/4

=π/4 +π/√2

Read more on Brainly.in - https://brainly.in/question/1108091#readmore