Question about finding the area of a function with delta x

[Juche]

I am new to this subject and I don't know what its officially called. I know that we are determining the area of a section of a graph below the graph of the function from points [a,b].

When I have a question that asks something like the area of the function 1/( x^2) on the interval 5,11 I just use my Ti-89 calculator and the 'ninc' function.

However one of my questions asks for the area on the interval 0,0.7 of delta x/(1-x^2)^(1/2). How do I figure this out exactly, this doesn't make sense to me. Shouldn't the area of the section under the function and above the x-axis be the height times width with delta x representing width and the function being height? My other questions were in the format function times delta x, I don't know how to do with with delta x divided by the function.

 

[vsage]

hrm what's wrong with \frac {\Delta x}{\sqrt{1-x^2}} ? if you take the function to be \frac {1}{\sqrt{1-x^2}} then you have it in function * delta x form.

 

[wais]

Finding the area under a curve is a fundamental concept in calculus. This process is known as integration, and it allows us to calculate the total area of a function between two given points. In this case, you are correct in thinking that the area should be the height times width, with delta x representing the width and the function representing the height. However, when calculating the area of a function that is divided by another function, we need to use a different method.

To find the area under the curve of a function divided by another function, we use a technique called integration by substitution. This involves substituting a new variable for the function in the denominator, and then using the chain rule to integrate the resulting function. In your case, you would substitute u=1-x^2, and then use the chain rule to integrate the function.

If you are new to this subject, it may be helpful to consult with your teacher or a tutor for further explanation and practice. Calculus can be a challenging subject, but with practice and understanding of the concepts, you will be able to solve problems like this one with ease. Keep up the good work and continue to ask questions and seek help when needed. Good luck!