How do you integrate $int 1/sqrt(2x+1)dx$ from [0,4]?

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Answer 1 · 2016-12-15T02:57:25

$2$ .

Let $u = (2x + 1)$ . Then $du = 2dx$ and $dx= (du)/2$

$=>1/2int_0^4(u^(-1/2)du)$

$=>1/2(2u^(1/2))|_0^4$

$=>u^(1/2)|_0^4$

$=>sqrt(2x+ 1)|_0^4$

We can now evaluate:

$=> sqrt(2(4) + 1) - sqrt(2(0) +1$

$=>sqrt(9) - sqrt(1)$

$=>3 - 1$

$=>2$

Hopefully this helps!

How do you integrate int 1/sqrt(2x+1)dxint 1/sqrt(2x+1)dx from [0,4]?