What is area f(x)= 1/x ,g(x)= 1/3 ,[2,5] ?

The answer to area f(x)= 1/x ,g(x)= 1/3 ,[2,5] is 2ln(3)+1/3-ln(10)

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area f(x)= 1/x ,g(x)= 1/3 ,[2,5]

area

f (x) = \frac{1}{x}, g (x) = \frac{1}{3}, [2, 5]

2 ln (3) + \frac{1}{3} - ln (10)

Decimal Notation

0.22797 \dots

Solution steps

Apply the area formula:

\int_{2}^{5} \frac{1}{x} - \frac{1}{3} d x

Solve

\int_{2}^{5} \frac{1}{x} - \frac{1}{3} d x : 2 ln (3) + \frac{1}{3} - ln (10)

The area is:

= 2 ln (3) + \frac{1}{3} - ln (10)

\frac{7 u ^{2}}{( u ^{2} + u ) ^{3}}

\int \frac{3 x ^{2}}{1 - x ^{6}} d x

x, x^{2} - 2

x \to 0 lim (3 cos (x))

\frac{x ^{2}}{2} - 4 x + 6, \frac{x}{2} - 1, [0, 5]

What is area f(x)= 1/x ,g(x)= 1/3 ,[2,5] ?
The answer to area f(x)= 1/x ,g(x)= 1/3 ,[2,5] is 2ln(3)+1/3-ln(10)