How to calculate this? int_0^3sqrtx/(sqrt(x)+sqrt(3-x))∫30√x√x+√3−x.

2 Answers
Cesareo R.
May 14, 2017

3/232

Explanation:

int_0^3sqrtx/(sqrt(x)+sqrt(3-x))=int_0^3(sqrtx (sqrt(x)-sqrt(3-x)))/(2x-3)=∫30√x√x+√3−x=∫30√x(√x−√3−x)2x−3=

int_0^3(x-sqrt(x(3-x)))/(2x-3) dx = 3/2∫30x−√x(3−x)2x−3dx=32

Ratnaker Mehta · mason m
May 14, 2017

3/2.32.

Explanation:

Using the Result, int_0^af(x)dx=int_0^af(a-x)dx,∫a0f(x)dx=∫a0f(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.!

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