Calculating Areas Between Curves

This video tutorial explains how to calculate the area between two curves using definite integrals. It covers the basic principles of integration for finding areas under curves and between curves, both with respect to x and y axes. The tutorial includes several practice problems, demonstrating the process of setting up integrals for different types of functions, such as linear and quadratic equations, and solving them to find the bounded area.

What is the integral used to find the area between two curves f(x) and g(x) from a to b?

• Integral from a to b of (f(x) - g(x)) dx

• Integral from a to b of (g(x) - f(x)) dx

• Integral from a to b of f(x) dx

• Integral from a to b of g(x) dx

When calculating the area between two curves in terms of y, which function is subtracted from which?

• Subtract the top function from the bottom function

• Subtract the bottom function from the top function

• Subtract the right function from the left function

• Subtract the left function from the right function

In the example of finding the area bounded by the line y = 8 - 2x, what are the x-intercepts?

• 2 and 8

• 4 and 8

• 0 and 4

• 0 and 8

How do you determine the points of intersection between y = x and y = x^2?

• Set y equal to x

• Set x equal to x^2

• Set y equal to x^2

• Set x equal to y

What is the area of the region bounded by y = x and y = x^2 from x = 0 to x = 1?

• 1/4

• 1/6

• 1/3

• 1/2

In the example with x = y^2 and y = x^2, what is the first step to find the area between the curves?

• Find the maximum value of each function

• Integrate each function separately

• Set the functions equal to each other

• Find the derivative of each function

What is the correct integral setup for finding the area between x = 1 - y^2 and x = y^2 - 1?

• Integral from -1 to 1 of (y^2 - 1) dy

• Integral from -1 to 1 of (1 - y^2 - (y^2 - 1)) dy

• Integral from -1 to 1 of ((y^2 - 1) - (1 - y^2)) dy

• Integral from -1 to 1 of (1 - y^2) dy

How do you find the vertex of the parabola y = x^2 - 4x?

• Use the formula x = -b/2a

• Find the midpoint of the x-intercepts

• Set the derivative equal to zero

• All of the above

What is the area of the region bounded by y = x^2 - 4x and y = 6 - 3x?

• 1/6

• 8/3

• 32/3

• 125/6

In the example with x = y^2 and x = 3y - 2, what is the first step to find the area between the curves?

• Find the derivative of each function

• Set the functions equal to each other

• Integrate each function separately

• Find the maximum value of each function