D. Applying for more credit cards
Answer:
i need help w this one to i’m doin it now
Explanation:
I have no clue what that is
Answer: 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](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B3%7D_%7B-3%7D%20%7Bx%5E2f%28x%5E3%29%7D%20%5C%2C%20dx%3D%20%5Cint%5Climits%5E%7B27%7D_%7B-27%7D%20%5Cfrac%7Bf%28u%29%7D3%20%5C%2C%20du%20%5C%5C%3D%5Cfrac%7B1%7D3%5Cleft%5B%5Cint%5Climits%5E%7B0%7D_%7B-27%7D%20%7Bf%28u%29%7D%20%5C%2C%20du%2B%5Cint%5Climits%5E%7B27%7D_%7B0%7D%20%7Bf%28u%29%7D%20%5C%2C%20du%20%5Cright%5D%20%282%29%5C%5CObserve%3A%5Cint%5Climits%5E%7B0%7D_%7B-27%7D%20%7Bf%28u%29%7D%20%5C%2C%20du%3D%5Cint%5Climits%5E%7B27%7D_%7B0%7D%20%7Bf%28u%29%7D%20du%5C%3B%20%5Ctext%7B%20because%20f%28x%29%20is%20an%20even%20function%7D%5C%5C%5Ctext%7BSubstitute%20the%20left%20hand%20side%20integral%20for%20the%20RHS%20in%20equation%20%282%29%3A%7D%5C%5C%3D%5Cfrac%7B1%7D3%5Cleft%5B2%5Cint%5Climits%5E%7B27%7D_0%20%7Bf%28u%29%7D%20du%5Cright%5D%5C%5C%3D%5Cfrac%7B1%7D3%20%282%29%2812%29%3D8)
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