What is area y= 1/x ,y= 1/(x^2),x=2 ?

The answer to area y= 1/x ,y= 1/(x^2),x=2 is ln(2)-1/2

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area y= 1/x ,y= 1/(x^2),x=2

area

y = \frac{1}{x}, y = \frac{1}{x ^{2}}, x = 2

ln (2) - \frac{1}{2}

Decimal

0.19314 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{1}^{2} \frac{1}{x} - \frac{1}{x ^{2}} d x : ln (2) - \frac{1}{2}

The area is:

= ln (2) - \frac{1}{2}

\frac{d}{d x} ((x - 5) (x + 4))

x^{3} + 4

\int \frac{1}{n n ^{2} - 1} d n

cos (e^{x^{5} c o s (x)})

\int_{6}^{8} \frac{96}{( x - 6 ) ^{3}} d x

What is area y= 1/x ,y= 1/(x^2),x=2 ?
The answer to area y= 1/x ,y= 1/(x^2),x=2 is ln(2)-1/2