Simple integration to find area

In summary, the conversation discusses finding the area under the curve y = 1/2^x between x=0 and x=1. The correct integration result is -1/(2^x ln2) + Const, which was confirmed on Wolfram. The area is calculated to be ~0.34 using the range x = 0 to x = 1. However, upon visual inspection of the curve, it is determined that the area should actually be double, or ~0.68. The range of the curve y = 1/2^x is from (0,1) to (1,0.5), and a rough visual calculation yields an approximate answer of 0.75. It is noted that
  • #1
Zman
96
0

Homework Statement


I wish to find the area under the curve y = 1/2^x between x=0 and x=1 but get an answer that is half the expected answer.

Homework Equations


Integrate y = 1/2^x to get -1/(2^x ln2) + Const

This integration result was confirmed on Wolfram

Slot in the range x = 0 to x = 1

Area = 1/ln2 (1 - 0.5) = ~0.34

The Attempt at a Solution


But when I look at the curve of y = 1/2^x

I can see that the area under the curve between x=0 and x = 1 is double

i.e. ~0.68The range of the curve y = 1/2^x that I am interested in starts at point (0,1) and goes to point (1, 0.5)

A rough visual calculation of this area is made up of the lower rectangle 1 x 0.5 added to the area of the upper triangle (1x0.5)/2 = 0.25 which gives an approximate answer of 0.75I seem to be missing a factor of 2 as regards the integral and am at a loss.
 
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  • #2
Zman said:
1/ln2 (1 - 0.5) = ~0.34
You seem to have calculated ln(2)/2, not 1/(2ln(2)).
 
  • #3
Thank you so much, this grey head is so greatful.
 

Related to Simple integration to find area

What is simple integration?

Simple integration is a mathematical technique used to find the area under a curve or between two curves on a graph.

Why is simple integration used to find area?

Simple integration is used because it allows for an accurate and precise calculation of the area under a curve, which may be difficult to do using other methods.

What are the steps involved in simple integration to find area?

The steps for simple integration to find area are:

  1. Identify the function that represents the curve or curves on the graph.
  2. Determine the limits of integration, or the points on the x-axis between which the area is to be calculated.
  3. Integrate the function with respect to x.
  4. Substitute the limits of integration into the integral and solve for the area.

What are some real-life applications of simple integration to find area?

Simple integration to find area is commonly used in physics, engineering, and economics to calculate the volume of irregular shapes, the work done by a variable force, and the area under a demand or supply curve, respectively.

Are there any limitations to using simple integration to find area?

Yes, there are limitations to using simple integration to find area. It can only be used for functions that have a well-defined antiderivative, and it is not applicable for finding the area of non-continuous or discontinuous functions.

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