Area Under Graph: Calculate (1+2^x) from 0 to 30

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In summary, you can take the area under a graph of (1+2^x) from 0 to 30 by doing the following: first, calculate the area under the curve y=1+2x. Then, add the area under the curve y=0 to the area under the curve y=1+2x. Finally, add the area under the curve y=30 to the area under the curve y=1+2x.
  • #1
whatdofisheat
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to take the area under a graph of
(1+2^x) from 0 to 30 how can you do this and what is the answer
 
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  • #2
[tex]\int\limits_0^{30} {\left( {1 + 2^x } \right)dx} = \left( {x + \frac{{2^x }}{{\log 2}}} \right)_0^{30} = 30 + \frac{{2^{30} - 1}}{{\log 2}}[/tex]

Questions like these belong here :)
 
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  • #3
If you have a function f(x), it's integral from a to b is the area bounded by the graph of the function, the x-axis and the lines y=a, y=b.

The term "area under the curve" can lead to confusion, because in the intervals where the function is negative, the integral actually gives the negative of the area not under but above the curve.

exemple: the integral of sin(x) between 0 and 2pi is 0 because there is a "mountain" over the x-axis, followed by a "valley" (under the x-axis). The total area under the mountain is A, and the area over the valley is A also, but since f(x) is negative there, this area is substracted while calculating the integral...

[tex]\int_0^{2\pi}\sin(x)dx = \int_0^{\pi}\sin(x)dx+\int_{\pi}^{2\pi}\sin(x)dx = A+(-A)=0[/tex]
 
  • #4
quasar987 said:
The term "area under the curve" can lead to confusion, because in the intervals where the function is negative, the integral actually gives the negative of the area not under but above the curve.
True indeed, but how does that apply here? 1+2x > 0 for all real x, so [itex]\int\limits_0^{30} {\left( {1 + 2x} \right)dx}[/itex] is (represents) indeed, as asked by the OP, the area under the curve y=1+2x (additionally bounded by y=0, x=0, and x=30).
whatdofisheat said:
to take the area under a graph of
(1+2^x) from 0 to 30 how can you do this and what is the answer
:redface: Hint:
[tex]\forall a > 0,\;\int {a^x dx} = \frac{{a^x }}{{\log a}} + C[/tex]
 
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  • #5
bomba923, when you delete a message do you have a "Physically remove" button below "Delete" and "Do Not Delete"? If so, clicking on that will not leave the "debris" behind that just using "Delete" does!

(I tried sending this as a private message but your box is full. Please go to your User CP and delete at least some of the private messages you have been getting.)
 
  • #6
HallsofIvy said:
bomba923, when you delete a message do you have a "Physically remove" button below "Delete" and "Do Not Delete"? If so, clicking on that will not leave the "debris" behind that just using "Delete" does!
I don't have a "physically remove" button anywhere! :frown:

*Can you add it to my account?
(*e.g., somewhat like adding avatar privileges to PF contributors' accounts, but in this case, adding a button/option around the "delete post" button...something like that, so I can physically my remove posts :redface:)
 

1. What is the purpose of calculating the area under the graph of (1+2^x) from 0 to 30?

The purpose of calculating the area under the graph is to determine the total area between the curve and the x-axis within the given interval. This can provide insights into the behavior and patterns of the function (1+2^x) within that interval.

2. How is the area under the graph of (1+2^x) from 0 to 30 calculated?

The area under the graph can be calculated using integral calculus. Specifically, the definite integral of (1+2^x) from 0 to 30 can be evaluated to find the total area under the curve within that interval.

3. What information can be obtained from the result of calculating the area under the graph?

The result of calculating the area under the graph can provide information such as the total change in the function (1+2^x) within the given interval, the average value of the function within that interval, and the behavior of the function in relation to the x-axis.

4. Are there any limitations to calculating the area under the graph of (1+2^x) from 0 to 30?

One limitation is that the calculation assumes the function (1+2^x) is continuous and well-behaved within the given interval. Additionally, the accuracy of the calculation may be affected by the method used to evaluate the integral.

5. How can the calculation of the area under the graph of (1+2^x) from 0 to 30 be applied in real-world scenarios?

The calculation of the area under the graph can be applied in various fields such as physics, economics, and engineering. For example, it can be used to calculate the work done by a varying force over a distance or the total revenue generated by a product over a specific time period.

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