# Determine whether the integral is convergent or divergent. integral_2^infinity 14/x^2 dx

## Question:

Determine whether the integral is convergent or divergent.

{eq}\displaystyle \int_2^{\infty} \frac{14}{x^2}\ dx {/eq}

## Integral:

We have to determine whether the integral is convergent or divergent.

Convergent means when we solve the integral and substitute the value of the limits, the limit comes to a finite value.

Divergent means when we solve the integral, the answer is not a finite value.

We have to determine whether the integral is convergent or divergent.

{eq}\displaystyle \int_2^{\infty} \frac{14}{x^2}\ dx {/eq}

Now:

{eq}\begin{align} \int_2^{\infty} \dfrac{14}{x^2}\ dx &=14\int_2^{\infty} \dfrac{1}{x^2}\ dx\\ &=14\int_2^{\infty} x^{-2}\ dx\\ &=14\left ( \dfrac{x^{-2+1}}{-2+1} \right )_2^{\infty}\ & \left [ \int x^ndx=\frac{x^{n+1}}{n+1} \right ]\\ &=14\left ( \dfrac{x^{-1}}{-1} \right )_2^{\infty}\\ &=-14\left ( \dfrac{1}{x} \right )_2^{\infty}\\ &=-14\left ( \dfrac{1}{\infty}-\dfrac{1}{2} \right )\\ &=-14\left ( 0-\dfrac{1}{2} \right )\\ &=-14\left ( \dfrac{-1}{2} \right )\\ &=7 \end{align} {/eq}

Hence the given integral converges.