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mandy9386

13.11.2022

Advanced Placement (AP)

A.) 8

Explanation:

Use u-substitution.

(1) Let u=x^3

By the power rule, du/dx=3x^2

Multiplying by dx and dividing by three, we have du/3=x^2dx

To find the new lower bound of integration, plug the old bound, -3, for x in equation (1). We get u=(-3)^3= -27

Similarly, when the upper bound 3 is plugged in, u=27

Now, replacing f(x^3) with f(u) and x^2dx with du/3:

$\int\limits^{3}_{-3} {x^2f(x^3)} \, dx= \int\limits^{27}_{-27} \frac{f(u)}3 \, du \\=\frac{1}3\left[\int\limits^{0}_{-27} {f(u)} \, du+\int\limits^{27}_{0} {f(u)} \, du \right] (2)\\Observe:\int\limits^{0}_{-27} {f(u)} \, du=\int\limits^{27}_{0} {f(u)} du\; \text{ because f(x) is an even function}\\\text{Substitute the left hand side integral for the RHS in equation (2):}\\=\frac{1}3\left[2\int\limits^{27}_0 {f(u)} du\right]\\=\frac{1}3 (2)(12)=8$

since the value of the first integral of the question = 12, which is given. Although the variable is different than the given (u instead of x), it's still the same integral

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