Find the area under the curve `y=1/(5x^3)` from x=1 to x=t and evaluate it for t=10, t=100. Then find the total area under this curve for `x>=1`
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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...
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