How do you integrate #int 1/sqrt(2x+1)dx# from [0,4]?
1 Answer
Dec 15, 2016
Explanation:
Let
#=>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!
