What is the net area between #f(x) = xsqrt(x^2-1) + 6# and the x-axis over #x in [2, 3 ]#?

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Answer 1 · 2017-01-09T18:48:44

#S= 6+(16sqrt(2))/3-sqrt(3)#

The are can be calculated solving the definite integral:

#S = int_2^3 (xsqrt(x^2-1)+6)dx = int_2^3 xsqrt(x^2-1)dx+6int_2^3dx#

Calculate the two integrals separately:

#6int_2^3dx = 6[x]_2^3 = 6#

#int_2^3 xsqrt(x^2-1)dx = 1/2int_2^3 sqrt(x^2-1)d(x^2-1) = 1/2[2/3(x^2-1)^(3/2)]_2^3 =1/3(8^(3/2)-3^(3/2))=(16sqrt(2))/3-sqrt(3)#

So:

#S= 6+(16sqrt(2))/3-sqrt(3)#