There's nothing stopping you integrating that, but it would not be ∫ydx. What is y equal to?Dumbledore211 said:I just can't integrate the function (a)^1/2-(x)^1/2 considering the fact that there are x as well as a
The limits should be at the axes, i.e. where x or y is zero. Does that correspond to x = 0 or a?should I take the limit from 0 to a for x?
Start by expanding the 'squared'.Dumbledore211 said:y={ (a)^1/2- (x)^1/2}^2. The integration should be done with respect to x where y is just a function of x. Can you drop a hint as to how I should integrate it and is 0≤ x≤ a or 0≤ x≤ (a)^1/2 true
I already answered that. The other bounds are the axes, so the limits correspond to x = 0 for one and y = 0 for the other.Dumbledore211 said:Okay, then it becomes y= a-2(a)^1/2(x)^1/2+x. Now, integrating both sides we get y= - 4/3x^3/2(a)^1/2+x^2/2. But, what limits should I take? Is 0≤x≤a valid in this case??
You missed a term when you integrated.Dumbledore211 said:Taking the limit from 0 to a the integration with respect to x becomes y= -4a^2/3+a^2/2= -8a^2+3a^2/6= -5a^2/6 How come I have a negative sign and the area is not negative. Besides the answer is a^2/6. What mistakes am i making?
No, the derivative of a constant would be zero, but not the integral.Dumbledore211 said:the integration of a becomes 0 with respect to x.
I don't see a second part. It asks you to find an area, which you have now done.Dumbledore211 said:Sorry, I made a ridiculously silly mistake. Don't know what happened as I solved harder integration problems like this. How do I solve the second part of the question?? I have no clue as to how I can find out the axes of the coordinates to be a^2/6
Let's not start a competition over who does the stupidest things. I might win.Dumbledore211 said:Sorry for my foolishness. I completely misread the question at first I thought that I had to show that the area and the axes of the coordinates are a^2/6. Anyway, Thank you haruspex for bearing my ridiculousness and showing me the extent of silliness that I exhibit in solving the simplest problems. My brain malfunctions at times especially when I am doing a lot of math per day...
The main purpose of using integration in scientific applications is to calculate the total or accumulated effect of a changing quantity over a given period of time or space. It allows us to find the area under a curve, which can have practical applications in fields such as physics, engineering, and economics.
Some common real-life examples of the application of integration include calculating the distance traveled by a moving object, finding the volume of a shape, determining the amount of work done by a variable force, and predicting the growth of a population over time.
Integration is used to solve differential equations by finding a function that satisfies the equation. By integrating both sides of the equation, we can eliminate the derivative and find the original function. This is a powerful tool in physics and engineering, where many natural phenomena can be described by differential equations.
Yes, integration can be used to find the average value of a function over a given interval. This is known as the mean value theorem and is calculated by finding the definite integral of the function over the interval and dividing by the length of the interval.
Integration is closely related to the concept of accumulation. Just as integration allows us to find the total effect of a changing quantity over a period of time, accumulation refers to the gradual increase or buildup of something over time. In essence, integration is a mathematical representation of the concept of accumulation.