Application of integration(the learning curve)

The production manager of an electronic company obtained the following function

$\displaystyle f(x)=1356.4x^{-0.3218}$

Where f(x) is the rate of labour hours required to assemble the $\displaystyle x^{th}$ unit of a product. The function is based on the experience of assembling the first 50 units of the product. The company was asked to bid on a new order of 100 additional units.

**find the total labour hours required to assemble 100 units **

SOLUTION: -

**N**=$\displaystyle \int_{50}^{150}f(x)dx=\int_{50}^{150}1356.4x^{-0.3218}$

**N**=$\displaystyle \left|\frac{1356.4x^{0.6782}}{0.6782}\right|_{50}^ {150}$

**N**=$\displaystyle 2000\left|150^{0.6782}-50^{0.6782}\right|$

Using logarithm and anti-logarithm,we get,

**N**=$\displaystyle 2000[29.91-14.2]$

**N**=31420

**Hence company can bid estimating the total labour hours needed 31420**

Verify this answer,

If I am wrong in finding the solution, reply me.

Re: Application of integration(the learning curve)

Hi Vinod! :)

Your reasoning is correct.

I do wonder though, whether you're supposed to make a continuity correction.

This would mean adding 0.5 to both integral boundaries.

Or is this out of scope of your material?

Btw, in your problem statement your have 1356, while in your calculation you use 1356.4.

This may be nitpicking, but are you supposed to do that?

Re: Application of integration(the learning curve)

Quote:

Originally Posted by

**ILikeSerena** Hi Vinod! :)

Your reasoning is correct.

I do wonder though, whether you're supposed to make a continuity correction.

This would mean adding 0.5 to both integral boundaries.

Or is this out of scope of your material?

Btw, in your problem statement your have 1356, while in your calculation you use 1356.4.

This may be nitpicking, but are you supposed to do that?

Hey,

Thanks for your reply.I have inadvertently typed in the problem statement '1356'instead of '1356.4' I am sorry about it. I have rectified my error in the problem now.