math problem (1 Viewer)

[Camsky]

Hey can someone help me answer this problem

calculate the area bounded by the graphs of f(x)=x(squared), g(x)=1over x(with the x squared), x>0 and the line x=3

this is question 3 fitzpatrick 2 unit page 367 for intergral calculus

thanks in advance

 

[life92]

f(x) = x^2
g(x) = 1/x^2

First, find the point of intersection.

x^2 = 1/x^2
x^4 = 1
Therefore x = 1 since x>0

If we draw a graph we can see that f(x)=x^2 > g(x) = 1/x^2 for x>1

Therefore the area =
S {3,1} x^2 - 1/x^2 dx
= [ x^3 / 3 + 1/x ] {3,1}
= (27 / 3 + 1/3) - (1/3 + 1)
= 8 units^2

 

[Camsky]

f(x) = x^2
g(x) = 1/x^2

First, find the point of intersection.

x^2 = 1/x^2
x^4 = 1
Therefore x = 1 since x>0

If we draw a graph we can see that f(x)=x^2 > g(x) = 1/x^2 for x>1

Therefore the area =
S {3,1} x^2 - 1/x^2 dx
= [ x^3 / 3 + 1/x ] {3,1}
= (27 / 3 + 1/3) - (1/3 + 1)
= 8 units^2

thank you very much for answering my question
that was perfectly explained