Question

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

Accepted Answer

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 = 0.09999; if t really large, area = 0.1

Math

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`

Expert Answers info

Unlock This Answer Now

Related Questions

Ask a Question

Popular Questions