Question

Найти площадь фигуры, ограниченной графиками функции y=x^2 +1 и y=10 (с полным решением)

Answer 1

X²+1=10⇒x²=9⇒x=+-3
S=$\int\limits {(10-x^2-1)} \, dx = \int\limits {(9-x^2)} \, dx$=9x-x³/3|(от -3 до 3)=
=27-9+27-9=54-18=36

Answer 2

Finding the Area of the Region Bounded by the Graphs of y=x^2+1 and y=10

To find the area of the region bounded by the graphs of the functions y=x^2+1 and y=10, we need to find the points of intersection between the two functions. The area can then be calculated by integrating the absolute difference between the two functions over the interval of intersection.

First, let's find the points of intersection by setting the two functions equal to each other and solving for x.

Step 1: Find the Points of Intersection The points of intersection can be found by setting the two functions equal to each other: x^2 + 1 = 10 x^2 = 9 x = ±3

So, the points of intersection are (-3, 10) and (3, 10).

Step 2: Calculate the Area The area can be calculated by integrating the absolute difference between the two functions over the interval of intersection: A = ∫[a, b] |f(x) - g(x)| dx

Where f(x) = x^2 + 1, g(x) = 10, and the interval [a, b] is [-3, 3].

Step 3: Evaluate the Integral A = ∫[-3, 3] |(x^2 + 1) - 10| dx A = ∫[-3, 3] |x^2 - 9| dx A = 2∫[0, 3] (x^2 - 9) dx (since the absolute value function is symmetric about the y-axis)

Now, let's calculate the integral to find the area.

Calculating the Integral

The integral can be calculated as follows: A = 2∫[0, 3] (x^2 - 9) dx A = 2 * [x^3/3 - 9x] evaluated from 0 to 3 A = 2 * [(3^3/3 - 9*3) - (0^3/3 - 9*0)] A = 2 * [(27/3 - 27) - (0 - 0)] A = 2 * [9 - 27] A = 2 * (-18) A = -36 square units

Conclusion

The area of the region bounded by the graphs of y=x^2+1 and y=10 is 36 square units.