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Svetach [21]
3 years ago
12

The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.

Mathematics
1 answer:
VLD [36.1K]3 years ago
3 0

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

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The first blanks options are -2 , -1/2 , 2 , 1/2
Pie

Answer:

equation 1

y = -2x

equation 2

y = x-3

Solution to both

(1,-2)

Step-by-step explanation:

We need to get the equations of both lines

General form is;

y = mx + c

where m is slope and c is the y-intercept

Table 1

since we have a point 0,0; the y-intercept here is zero

Let us get the slope. We can do this by selecting any two points

m = (y2-y1)/(x2-x1)

m = (2-10)/(-1+5) = -8/4 = -2

So the equation of the first line is;

y = -2x

Table 2

we get the slope

m = (4+2)/(7-1) = 6/6 = 1

The partial equation is;

y = x + c

To get c, we select any two point and substitute

4 = 7 + c

c = 4-7

c = -3

So the equation is;

y = x-3

To get the solution to both systems, we equate the y

-2x = x - 3

-2x-x = -3

-3x = -3

x = -3/-3

x = 1

To get y, we substitute;

recall; y = -2x

y = -2(1)

y = -2

Solution to the system is;

(1,-2)

7 0
3 years ago
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