What is the net area between f(x) = x^2-sqrt(x+1) and the x-axis over x in [1, 7 ]?

1 Answer
Dec 14, 2017

See below.

Explanation:

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:

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