Question

обчислити площу фігури, обмежену лініями:
у= -х^2+8х
у=х^2+18х-12
​помогите​

Answer 1

Ответ:

$y=-x^2+8x\ \ ,\ \ y=x^2+18x-12\\\\x^2+18x-12=-x^2+8x\ \ ,\ \ 2x^2+10x-12=0\ \ ,\ \ \ x^2+5x-6=0\ ,\\\\x_1=-6\ ,\ x_2=1\ \ \ (teorema\ Vieta)\\\\\\\displaystyle S=-\int\limits^1_{-6}\, (-x^2+8x-x^2-18x-12)\, dx=-\int\limits^1_{-6}\, (-2x^2-10x-12)\, dx=\\\\\\=-\Big(-\frac{2x^3}{3}-5x^2-12x\Big)\Big|_{-6}^1=-\Big(-\frac{2}{3}-5-12\Big)+\Big(\frac{2\cdot 6^3}{3}-5\cdot 6^2+12\cdot 6\Big)=\\\\\\=17+\frac{2}{3}-108+\frac{432}{3}=-91+\frac{434}{3}=\frac{161}{3}$