Find the area of the region bounded by the curves.

Question

Directions: Find the area of the region bounded by the curves. Use a clean sheet of paper for your solutions and final answers.

questions: https://imgur.com/y718s83

Answer 1

It is helpful to graph the function and the given lines to visualize the enclosed area between them. Desmos calculator can help you with the graph. The link is desmos.com/calculator . It is important to remember that the enclosed area that is above the x-axis will give you a positive integral value and the enclosed area that is below the x-axis will give you a negative integral value. In that last case, since this is an area and the area must be positive you can put a negative sign before the integral sign that converts that negative value into a positive value. Since there 5 exercises I will help you to set the integral but you have to work them out using the integral rules and properties.

1) If you graph y = 4 - x^2 you will see that the intersection between the parabola and the x-axis is in x = -2 and x = 2 and the enclosed area is above the x-axis. So area would be given by the definite integral from -2 to 2 of (4 - x^2) dx. The answer will be 32/3.

2) If you graph y = x^2 - 2x + 3, and the lines x = -2 and x = 1 together with the x-axis you will see that the enclosed area is above the x-axis. Here the x-values for the interval where the enclosed area is, are given. So area would be given by the definite integral from -2 to 1 of (x^2 - 2x +3) dx. The answer will be 15.

3) If you graph y = √(x + 1), and the line x = 8 together with the x-axis and y-axis you will see that the enclosed area is above the x-axis. Here the x-values for the interval where the enclosed area will be x = 0 and x = 8. So area would be given by the definite integral from 0 to 8 of √(x + 1) dx. The answer will be 52/3.

4) First solve the equation x^2 = -y for y and you will get y = -x^2. This exercise is not using the x-axis like the first three. So the solution is worked differently. If you graph y = -x^2 and y = -4 you will see y = -x^2 is above y = -4 and that the intersection between them is in x = -2 and x = 2. So when setting the integral will be a difference between the above function and the below function. The area would be given by the definite integral from -2 to 2 of (-x^2 - (-4)) dx. The answer is 32 / 3.

5) If you graph y = sin x, and the lines x = (1/3)π and x = (2/3)π together with the x-axis you will see that the enclosed area is above the x-axis. Here the x-values for the interval where the enclosed area will be x = (1/3)π and x = (2/3)π. So area would be given by the definite integral from (1/3)π to (2/3)π of sin x dx. The answer will be 1.