Answer:
I need help on number 29 please explain this too.
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If you look at the graph below, the "ceiling" function, is f(x), on that interval of [-4,0], whilst the "floor" function, is g(x)
so, taking that into account, then![\bf \begin{cases} f(x)=x^3+x^2-6x\\\\ g(x)=6x \end{cases}\qquad \displaystyle \int\limits_{-4}^{0}\ \begin{array}{llll} [ceiling]&-&[floor]\\ f(x)&&g(x) \end{array} \\\\\\ thus \\\\\\ \displaystyle \int\limits_{-4}^{0}\ [[x^3+x^2-6x]-[6x]]dx\implies \displaystyle \int\limits_{-4}^{0}\ x^3+x^2-12x \\\\\\ \left. \cfrac{x^4}{4}+\cfrac{x^3}{3}-6x^2 \right]_{0}^{-4}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Af%28x%29%3Dx%5E3%2Bx%5E2-6x%5C%5C%5C%5C%0Ag%28x%29%3D6x%0A%5Cend%7Bcases%7D%5Cqquad%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Bceiling%5D%26-%26%5Bfloor%5D%5C%5C%0Af%28x%29%26%26g%28x%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%0A%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20%5B%5Bx%5E3%2Bx%5E2-6x%5D-%5B6x%5D%5Ddx%5Cimplies%20%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20x%5E3%2Bx%5E2-12x%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%20%5Ccfrac%7Bx%5E4%7D%7B4%7D%2B%5Ccfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%20%20%5Cright%5D_%7B0%7D%5E%7B-4%7D)
so, taking that into account, then
Answer:
Step-by-step explanation:
Let f(y)=x