hmmm... I'm very low on English but if i understood correctly, u need to change limits of integration... Isn't that for ?
Just wondering if there's at the end
This problem involves a change in variable of the limit of integration.
Consider the area bounded by , y=0, x=0, and x=.5. A definite integral of this area, using x as the variable of integration, is
. The next part of the question says: make the substitution of the form to transform the integral so that is the variable of integration.
I'm not sure how to start this one. I thought of simply solving for , but I'm don't know how to transform the integrand.
*edit* forgot my dx. x.x
I would assume you want to evaluate
The key is to think of it like a triangle(check triangle.pdf)
And lastly you have
Now the last part is to change the limits of integration. You have so and
Can you finish from here?
You don't need to change the limits of integration if you don't want to. You can write it like this and then change back to x when you've solved the integral before plugging in the numerical values and calculating the difference.
Okay, I get that, and was able to write out my definite integral. The next part of the question asks to evaluate each of the integrals. Should I be getting the same or different answers for the two integrals? I got .40361 for the first answer, and .697067 for the second answer.
Yes you should be getting the same answers. Here I'll show you a simpler example. Consider the integral
In your triangle put the hypotenuse equal to 1. The leg that is across the angle should be x so that
Now using Pythagoras theorem the other leg is
Now you can write the integral like this
You have to express in terms of
We have and we differentiate both sides to get
Now we substitute dx in the integral to get
What you do next is change the limits of integration. The limits are x=0 and x=1 but we are integrating with respect to
You use and you need to solve the two equations.
You have to work in radians. The reason for that is because radians have a unit of length which is the same unit for x and y.
Now your integral becomes