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Definite Integrals
This notebook will help you with the following:
- Estimating area under curves using left and right endpoint approximations.
- Finding area under postive curves using definite integrals, as well as finding diplacement from velocity.
- Finding area between functions and the x axis.
- Finding average value.
Approximating Area
If you enter a function, a range of values from to , and a number of rectangles you want to create, the code below will compute both left and right endpoint approximations to the directed area under . I said directed because if , then you will get a negative, and if then you will get a negative as well.
x | 0 | \frac{1}{2} | 1 | \frac{3}{2} | 2 |
y | 4 | \frac{15}{4} | 3 | \frac{7}{4} | 0 |
dA = y dx | 2 | \frac{15}{8} | \frac{3}{2} | \frac{7}{8} | 0 |
Left Sum | \frac{25}{4} | 6.25000000000000 |
Right Sum | \frac{17}{4} | 4.25000000000000 |
Average Sum | \frac{21}{4} | 5.25000000000000 |
Actual Area | \frac{16}{3} | 5.33333333333333 |
Definite Integrals
The following code will compute definite integrals, as well as graph a function and shade the region between the function and the -axis.
- If you type in velocity, it will find the change in position.
- If you type in a positive function, it will compute the area under the graph.
You can type in your functions using , , or any other variable you wish.
Area between and the -axis.
If you want to find the area between and the -axis, the first step is to find the zeros of the function. Then you break the integral to integrate over positive and negative parts separately. The code below will find the zeros, give you the integral, tell you the area from both the positive and negative parts, as well as the total area.
x intercepts | \text{[ x == 3, x == -2 ]} |
Integral of f | \frac{1}{3} \, x^{3} - \frac{1}{2} \, x^{2} - 6 \, x |
Total Area | 65.0 |
Positive Area | 44.1666666667 |
Negative Area | 20.8333333333 |
Average Value
The code below computes the average value of a function. Remember that the average value of a continuous function always passes through the function. The average value is the value so that the area above below is the same as the area below and above . I like to think of average value as the height of sand in an ant farm that would result from shaking the ant farm to make a level top.
Average Value | \frac{8}{3} |
x values | \text{[ x == -2/3*sqrt(3), x == 2/3*sqrt(3) ]} |
Notice how the area above the average value is the same as the area below. |