The average value of a function $f(x)$ on an interval $[a, b]$ is given by the formula:

$f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \,dx$

Given function:

$f(x) = \frac{x^2}{(x^3 + 7)^2}$

Given interval:

$[-1,1]$

Thus, we compute:

$f_{\text{ave}} = \frac{1}{1 - (-1)} \int_{-1}^{1} \frac{x^2}{(x^3 + 7)^2} \,dx$

This simplifies to:

$f_{\text{ave}} = \frac{1}{2} \int_{-1}^{1} \frac{x^2}{(x^3 + 7)^2} \,dx$

Substitution:

Let $u = x^3 + 7$, then differentiating both sides:

$du = 3x^2 dx \quad \Rightarrow \quad \frac{du}{3} = x^2 dx$

Rewriting the integral in terms of $u$:

$\int_{-1}^{1} \frac{x^2}{(x^3 + 7)^2} \,dx = \int_{-1}^{1} \frac{x^2}{u^2} \,dx$

Using the substitution:

$\int_{u(-1)}^{u(1)} \frac{\frac{du}{3}}{u^2}$

$\frac{1}{3} \int_{u(-1)}^{u(1)} u^{-2} \,du$

$\frac{1}{3} \left[ -\frac{1}{u} \right]_{u(-1)}^{u(1)}$

Evaluating at $x = -1 \Rightarrow u(-1) = (-1)^3 + 7 = 6$, and at $x = 1 \Rightarrow u(1) = 1^3 + 7 = 8$:

$\frac{1}{3} \left[ -\frac{1}{8} + \frac{1}{6} \right]$

Finding a common denominator (24):

$\frac{1}{3} \left[ -\frac{3}{24} + \frac{4}{24} \right] = \frac{1}{3} \times \frac{1}{24} = \frac{1}{72}$

Now, multiply by $\frac{1}{2}$ for the average value:

$f_{\text{ave}} = \frac{1}{2} \times \frac{1}{72} = \frac{1}{144}$

Final Answer:

$f_{\text{ave}} = \frac{1}{144}$

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Related Questions:

1. How do you find the average value of a function on any given interval?
2. What is the significance of the average value of a function?
3. How does substitution simplify complex integrals?
4. Why is the function symmetric over the interval $[-1,1]$?
5. What are common mistakes when evaluating definite integrals?

Tip:

When using substitution in integrals, always remember to adjust the limits of integration accordingly to avoid back-substituting at the end.