Question

Обчисліть площу фігури, обмежену графіками функцій y=2x+6 i y=(x+3)^2 Виконайте рисунок.

Answer 1

Объяснение:

$y=2x+6\ \ \ \ \ y=(x+3)^2\ \ \ \ \ S=?\\2x+6=(x+3)^2\\2x+6=x^2+6x+9\\x^2+4x+3=0\\D=4\ \ \ \ \ \sqrt{D}=2\\ x_1=-3\ \ \ \ x_2=-1.\\S=\int\limits^{-1}_{-3} {(2x+6-(x+3)^2)} \, dx=\int\limits^{-1}_{-3} {(2x+6-x^2-6x-9))} \, dx=\\=\int\limits^{-1}_{-3} {(-4x-3-x^2)} \, dx=-\int\limits^{-1}_{-3} {(x^2+4x+3)} \, dx =-(\frac{x^3}{3}+2x^2+3x)| _{-3}^{-1}=\\=-((\frac{(-1)^3}{3}+2*(-1)^2+3*( -1)-(\frac{(-3)^3}{3}+2*(-3)^2+3*(-3)))=\\ -(-\frac{1}{3} +2-3-(-9+18-9))=-(-1\frac{1}{3} -0)=\frac{4}{3} .$

Ответ: S=1,33333 кв. ед.

Answer 2

Calculating the area of the figure bounded by the graphs of functions y=2x+6 and y=(x+3)2

To calculate the area of the figure bounded by the graphs of the given functions, we first need to draw the graph. The graph of the function y=2x+6 is a line with slope 2 that passes through the point (0,6). The graph of the function y=(x+3)2 is a parabola that opens upwards and passes through the point (-3,0).

We can see that the intersection points of the two graphs are (-3,0) and (3,12). These points give us the boundaries of the area we need to calculate.

To find the area, we calculate the definite integral of the region bounded by the functions. The formula for the area bounded above by f(x) and below by g(x) between the points x=a and x=b is:

Area = ∫from a to b [f(x) - g(x)]dx

Plugging in the functions and limits of integration, we get:

Area = ∫from -3 to 3 [(x+3)2 - 2x - 6]dx = ∫from -3 to 3 [x2 + 6x + 9 - 2x - 6]dx = ∫from -3 to 3 [x2 + 4x + 3]dx = [x3/3 + 2x2 + 3x] from -3 to 3 = 36

Therefore, the area of the region bounded by the graphs of y=2x+6 and y=(x+3)2 is 36 units2.