How do you find the definite integral of $int sqrt(x) dx$ from $[0,3]$ ?

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Answer 1 · 2015-12-11T12:50:15

$int_0^3 sqrt(x).dx= 2 sqrt(3)$

Been quite a while since I did any Integration so hopefully I am correct!

Let $y=int_0^3 sqrt(x).dx$

Write as : $y=int_0^3 x^(1/2).dx$

note that $int x^a.dx =1/(a+1) x^(a+1) + C$

$y=[ 2/3x^(3/2)+C]_0^3$

$y= [ 2/3(3)^(3/2)] -[ 2/3(0)^(3/2)] +(C-C)$

$y=2/3 xx sqrt(27)$

But $27=3xx3^2$

so $y= 2/3xx 3sqrt(3)$

$y= 2 sqrt(3)$

How do you find the definite integral of int sqrt(x) dxint sqrt(x) dx from [0,3][0,3]?