What is area f(x)= 1/x ,g(x)=-1/3 x+3,[0,12] ?

The answer to area f(x)= 1/x ,g(x)=-1/3 x+3,[0,12] is diverges

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area f(x)= 1/x ,g(x)=-1/3 x+3,[0,12]

area

f (x) = \frac{1}{x}, g (x) = - \frac{1}{3} x + 3, [0, 12]

diverges

Solution steps

Apply the area formula:

\int_{0}^{12} \frac{1}{x} - (- \frac{1}{3} x + 3) d x

Solve

\int_{0}^{12} \frac{1}{x} - (- \frac{1}{3} x + 3) d x :

diverges

The area is:

= diverges

\int_{0}^{1} \frac{4 x}{e ^{x^{2}}} d x

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{0.5})

\int \frac{4 - x}{x ^{4} - x ^{2}} d x

x \to 2 lim (2 x + c)

f (x) = \frac{1 - ln ( x )}{2 x ^{2}}

What is area f(x)= 1/x ,g(x)=-1/3 x+3,[0,12] ?
The answer to area f(x)= 1/x ,g(x)=-1/3 x+3,[0,12] is diverges