What is the integral from 2 to infinity of 1/(x^2) ?

The integral from 2 to infinity of 1/(x^2) is 1/2

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integral from 2 to infinity of 1/(x^2)

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{x ^{2}} d x = - \frac{1}{x} + C

Compute the boundaries:

\int_{2}^{\infty} \frac{1}{x ^{2}} d x = 0 - (- \frac{1}{2})

= 0 - (- \frac{1}{2})

Simplify

= \frac{1}{2}

\int_{1}^{3} (3 x^{2} - 2 x + 1) d x

\int 16 sin^{2} (θ) d θ

f (x) = \frac{A}{x ^{14}} + B e^{x}

y = 2 + x - 7

x \to \infty lim (e^{\frac{1}{x} l n (2^{x} + 3^{x} + 5^{x})})

What is the integral from 2 to infinity of 1/(x^2) ?
The integral from 2 to infinity of 1/(x^2) is 1/2