How to calculate this? #int_0^3sqrtx/(sqrt(x)+sqrt(3-x))#.

Question

1007 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-14T10:24:58

#3/2#

#int_0^3sqrtx/(sqrt(x)+sqrt(3-x))=int_0^3(sqrtx (sqrt(x)-sqrt(3-x)))/(2x-3)=#

#int_0^3(x-sqrt(x(3-x)))/(2x-3) dx = 3/2#

Answer 2 · 2017-05-14T11:49:06

# 3/2.#

Using the Result, #int_0^af(x)dx=int_0^af(a-x)dx,# we have, for,

#I=int_0^3sqrtx/{sqrtx+sqrt(3-x)}dx.................(1).#

# I=int_0^3sqrt(3-x)/{sqrt(3-x)+sqrt(3-(3-x))}dx, i.e., #

# I=int_0^3sqrt(3-x)/{sqrt(3-x)+sqrtx}dx................(2).#

Adding #(1) and (2),# we have,

#I+I=2I=int_0^3{sqrtx+sqrt(3-x)}/{sqrtx+sqrt(3-x)}dx.#

# rArr 2I=int_0^3 1dx=[x]_0^3=3-0=3.#

# :. I=3/2#, as Respected Cesareo R., has readily obtained!

Enjoy Maths.!