The marginal cost of manufacturing x yards of a certain fabric is C'(x)=3-0.01x+0.000006x^2 ( in dollars per yard ). Find the increase in cost if the production level is raised from 500 yards to 5000 yards.

and

fnInt((x^3 +3 + 1/(x^2+1))dx)

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- Jun 11th 2007, 04:49 PMsw3etazngyrlIntegrals
The marginal cost of manufacturing x yards of a certain fabric is C'(x)=3-0.01x+0.000006x^2 ( in dollars per yard ). Find the increase in cost if the production level is raised from 500 yards to 5000 yards.

and

fnInt((x^3 +3 + 1/(x^2+1))dx) - Jun 11th 2007, 05:02 PMJhevon
I suppose for this one you want:

$\displaystyle \int_{500}^{5000} \left( 3 - 0.01x + 0.000006x^2 \right)dx$

that is not a hard function to integrate. just use the power rule. unless you should be calculating something else. the only place i encountered things like "marginal cost" and such was in a microeconomics class, and we never used calculus at all, so i'm not sure how to relate everything

Quote:

and

fnInt((x^3 +3 + 1/(x^2+1))dx)

- Jun 11th 2007, 05:17 PMsw3etazngyrl
fnInt means integral

- Jun 11th 2007, 05:22 PMJhevon
ok. again, most of this integral is simple, you use the power rule. but you also have to know something else.

Power Rule: $\displaystyle \int x^n dx = \frac {x^{n+1}}{n+ 1} + C$ for $\displaystyle n \neq -1$

Something else: $\displaystyle \int \frac {1}{x^2 + 1}dx = \arctan x + C$

So,

$\displaystyle \int \left(x^3 + 3 + \frac {1}{x^2 + 1} \right)dx = \frac {1}{4}x^4 + 3x + \arctan x + C$

Now, can you do the first?