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)#