Solving the integral?

Question

The integral from 0 to 1 of squareroot(2+4y^2) dy. Can anyone help me solve this problem please.

Thank you!!!

Answer 1

Int[ sqrt(2+4y^2)dy ] (Limits are plugged after solving indefinite integral)
=Int[ sqrt(4 ( (1/2)+y^2 ) )dy
=2Int[ sqrt( (1/2)+y^2 )dy
=2[ ( y/2).sqrt( (1/2)+y^2 ) + (1/4) log ly + sqrt( (1/2)+y^2 )l ] + C (Apply formula Int[ sqrt(a^2 + x^2) dx ]
=2[(1/2)sqrt( (1/2) + 1 ) + (1/4) log l1 +sqrt( (1/2)+1)] - 2[ (1/4) log lsqrt(1/2)l ] (Plug in the limits)
=sqrt(3/2) + [ log lsqrt(2) + sqrt(3)l ]/2 + C
Welcome!

Answer 2

For the indefinite integral sqrt(a^2+y^2)dy the formula is
(y/2)*sqrt(a^2+y^2) + [(a^2)/2]*ln[y+sqrt(y^2+a^2)]

in sqrt(2+4y^2) take 4 out of the root then we get
2*sqrt(1/2 + y^2)
here a^2=1/2
and we have y^2
substituting in the formula we get an indefinite integral where on substituting the limits 0 & 1 we get the required answer