x-sqrt(x+1)=0
x^2-x-1=0=>x=(1+sqrt(5))/2 and x=(1-sqrt(5))/2
The function is below the x axis in the interval:
((1-sqrt(5))/2, (1+sqrt(5))/2)
So we need two integrals:
A=-int_(1)^((1+sqrt(5))/2)(x^2-sqrt(x+1)) dx+int_((1+sqrt(5))/2)^(7)(x^2-sqrt(x+1)) dx
A=-int_(1)^((1+sqrt(5))/2)(x^2-sqrt(x+1)) dx
->=[1/3x^3-2/3(x+1)^(3/2)]^((1+sqrt(5)))-[1/3x^3-2/3(x+1)^(3/2)]_(1)
=-[1/3((1+sqrt(5))/2)^3-2/3((1+sqrt(5))/2+1)^(3/2)]-[1/3(1)^3-2/3((1+sqrt(5))/2+1)^(3/2)]=0.04738
A=int_((1+sqrt(5))/2)^(7)(x^2-sqrt(x+1)) dx
->=[1/3x^3-2/3(x+1)^(3/2)]^(7)-[1/3x^3-2/3(x+1)^(3/2)]_((1+sqrt(5))/2)
->=[1/3(7)^3-2/3(7+1)^(3/2)]^(7)-[1/3((1+sqrt(5))/2)^3-2/3(((1+sqrt(5))/2)+1)^(3/2)]_((1+sqrt(5))/2)=100.84806
Area=100.84806+0.04738=100.89 units squared.
Graph: