How do I work this Calculus problem?!?

Hi Jared.

A) Find the area of "R".

This is going to require integrals. You need to take the integral from "0" to "√6" of the equation given.

∫ [x / (x² + 2)]dx

Take the antiderivative F(x) of the integral.

F(x) = ln(x² + 2) / 2

Take the sum of the antiderivative on the interval given.

[ln(√6² + 2) / 2] - [ln(0² + 2) / 2]

Simplify.

[ln(8) / 2] - [ln(2) / 2]

Simplify further.

[ln(2³) / 2] - [ln(2) / 2]

Simplify further.

[3ln(2) / 2] - [ln(2) / 2]

Simplify further.

2ln(2) / 2

Simplify further.

ln(2) <= FINAL ANSWER

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Unfortunately we haven't been taught how to do this in class yet, so I can't help you with this one.

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C) What is the average value of the function on its interval?

For this, you need to use the Mean Value Theorem (for integrals). This theorem states the following:

If f is integrable on [a, b], then its mean value on that interval is

[1 / (b - a)] × [the integral from "a" to "b" of f(x)].

So basically just follow the equation that they give you and substitute in your values. We already found the integral of the function, so really you are just multiplying it by a fraction.

[1 / (√6 - 0)] × ln(2)

Simplify.

1 / (√6) × ln(2)

Simplify further.

ln(2) / (√6)

Rationalize the denominator.

(√6)ln(2) / 6 <= FINAL ANSWER

I hope this helped. (:

f (x) = x^3 - 12x + a million . . . the first by-product set to 0 unearths turning or table certain factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the second one by-product evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== useful fee shows x=2 is an section minimum f ' ' (-2) = 6*(-2) = -12 <== damaging fee shows x=-2 is an section optimal a.) x = - 2 is a optimal, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is lowering x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the second one by-product set to 0 unearths inflection factors, or the position concavity alterations 6x = 0 x = 0 <=== inflection element x = - 2 is a optimal, so must be concave down concavity alterations on the inflection element(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up