We have `y = 1/(5x^3)`

`int_1^t 1/(5x^3) dx = -1/(2(5x^2))|_1^t` (raise the power by one and divide by the new power)

` = - 1/(2(5t^2)) - (- 1/(2(5(1^2)))) = -1/(2(5t^2)) + 1/10`

Evaluating for t = 10 gives ` - 1/(2(5(10^2))) + 1/10 = -1/(2(5(100))) + 1/10 = -1/1000 +...

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We have `y = 1/(5x^3)`

`int_1^t 1/(5x^3) dx = -1/(2(5x^2))|_1^t` (raise the power by one and divide by the new power)

` = - 1/(2(5t^2)) - (- 1/(2(5(1^2)))) = -1/(2(5t^2)) + 1/10`

Evaluating for t = 10 gives ` - 1/(2(5(10^2))) + 1/10 = -1/(2(5(100))) + 1/10 = -1/1000 + 1/10 = -0.001 + 0.1 = 0.099`

Evaluating for t = 100 gives

`- 1/(2(5(100^2))) + 1/10 = - 1/(2(5(10000))) + 1/10 = -1/100000 + 1/10 `

`= -0.00001 + 0.1 = 0.09999`

The total area under the curve for x >= 1 is the area from x = 1 to `lim_(t->oo) (t)` ` `

From what we have already done we can see that as t gets larger and larger the area under the curve tends to 0.1

So in algebraic notation

`int_1^oo 1/(5x^3) dx = - 1/(2(5x^2))|_1^t = 1/10 - lim_(t->oo) (1/(2(5t^2))) = 1/10 - 0 = 1/10`

**If t = 10, area = 0.099; if t = 100, area = 0.0999****9; if t really large, area = 0.1**