Question

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

Answer 1

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

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 = (x^2)/(2)`

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

`\int\limits^{}_{} {x^2} \, dx = (x^3)/(3)`

Then our integral is:

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