area under the curve
TRANSCRIPT
![Page 1: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/1.jpg)
![Page 2: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/2.jpg)
What is “area under the curve?”
• The definite integral can be used to find the area between a graph curve and the ‘x’ axis, between two given ‘x’ values. This area is called the ‘area under the curve’ regardless of whether it is above or below the ‘x’ axis.
![Page 3: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/3.jpg)
• When the area is above the x axis, the area is positive.
• When the area is below the x axis, the area is negative.
![Page 4: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/4.jpg)
Example problem
• y = x² - x – 2• We need to create three
separate integrals
![Page 5: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/5.jpg)
Finding the integrals
• The zeros of the function f(x) that lie between -2 and 3 form the boundaries of the separate area segments, and these will be our integrals
• -2 < x < -1 • -1 < x < 2• 2 < x < 3
![Page 6: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/6.jpg)
In Integral Form:
![Page 7: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/7.jpg)
• Integrate the equation using the general power rule:
• 1/3x^3 – 1/2 x^2 – 2x
![Page 8: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/8.jpg)
• Plug in beginning points and endpoints for each of three intervals and subtract the beginning point from the end point.
![Page 9: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/9.jpg)
Plug in numbers and solve
• 1/3(-1)^3 – ½ (-1)^2 – 2(-1) – (1/3(-2)^3 – ½ (-2)^2 – 2(-2) = 1.83333
• 1/3(2)^3 – ½ (2)^2 – 2(2) – (1/3(-1)^3 – ½ (-1)^2 – 2(-1)) = 4.5
• 1/3(3)^3 – ½ (3)^2 – 2(3) – (1/3(2)^3 – ½ (2)^2 – 2(2)) = 1.83333
![Page 10: Area Under the Curve](https://reader035.vdocuments.site/reader035/viewer/2022062418/5559c6edd8b42aaa6f8b542e/html5/thumbnails/10.jpg)
• The total shaded area will be A = A1 + A2 + A3
• Total area = 1.83333 + 4.5 + 1.83333 = 8.16666
• The net area takes into account the negative area:
• 1.83333 - 4.5 + 1.83333 • = - 0.83333