# Evaluate the integral: integral_{1}^{infinity} (3) / (x^6) dx.

## Question:

Evaluate the integral:

{eq}\displaystyle \int_{1}^{\infty} \frac{3}{x^6}\ dx {/eq}.

## Improper Integrals:

Improper Integrals contains one or both of the limits of integration as infinity. Such an integral is often written symbolically just like a standard definite integral, in some cases with infinity as a limit of integration:

{eq}\displaystyle{ \int _ { a } ^ { \infty } f ( x ) d x = \operatorname { lim } _ { t \rightarrow \infty } \int _ { a } ^ { t } f ( x ) d x }{/eq}

provided the limit exists and is finite.

We have,

{eq}\displaystyle{ \displaystyle \int_{1}^{\infty} \frac{3}{x^6}\ dx }{/eq}

We can rewrite this as:

{eq}\displaystyle{ \operatorname { lim } _ { m \rightarrow \infty } \int _ { 1 } ^ { m } \frac { 3 } { x ^ { 6 } } d x }{/eq}

where {eq}\displaystyle{ m }{/eq} approaches infinity.

Apply power rule of integration {eq}\displaystyle{ \int x ^ { n } d x = \frac { x ^ { n + 1 } } { n + 1 } + C }{/eq}, we get:

{eq}\displaystyle{ \operatorname { lim } _ { m \rightarrow \infty } \int _ { 1 } ^ { m } 3 x ^ { - 6 } d x\\ =\operatorname { lim } _ { m \rightarrow \infty }3 \frac { [ x ^ { - 5 } ] _ { 1 } ^ { m } } { - 5 }\\=- \frac { 3 } { 5 } \operatorname { lim } _ { m \rightarrow \infty }( m ^ { - 5 } - 1 )\\= - \frac { 3 } { 5 } \operatorname { lim } _ { m \rightarrow \infty } ( \frac { 1 } { m ^ { 5 } } - 1 )\\ = - \frac { 3 } { 5 } ( \frac { 1 } { \infty } - 1 ) = \frac { 3 } { 5 } }{/eq}

Hence, {eq}\boxed{ \displaystyle \displaystyle \int_{1}^{\infty} \frac{3}{x^6}\ dx=\frac { 3 } { 5 } }{/eq}