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

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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.!

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