Help with evaluating integrals?

[khadija112]

I started with negating the integral and switching the boundaries (since 1 is less than 2) then took the anti derivative. Afterwards I plugged in one for the x value and got -1/3 -3/2. I’m pretty sure I’m doing something wrong but I don’t know what.. my attempt at this problem was a guess anyways so I’m very unsure of what steps to take. Please help lol

 

[Dr.Peterson]

You didn't really need to change the order of the limits, but it's not wrong. You were right up to a point.

But why didn't you evaluate the upper limit?

​

And why didn't you add the fractions?

I hope you see that g(2) is easy. If it weren't for that, I would strongly recommend finding the general form of g(x) before using it to find g(1).

 

[khadija112]

We aren’t required to simplify for responses so I usually leave my work alone. I found the answer to g(2) and got it correct but for some reason I was overthinking the first portion. Thank you for your advice

 

[Steven G]

I have a few comments on your work/comments.

1) Did you draw the graph and see that the graph is above the x-axis between x=1 and x=2? If not, then how do you know that the area/answer should be positive? Or are all integrals supposed to be positive.

2) Yes, you are correct that if you interchange the limits of integration you must factor out a -1. My question is why would you want to do that given that in your case it makes the evaluation so awkward? Sure, if you already have a negative number factored out in front of the integral , then interchanging the limits will cause you do factor out a -1 making the number factored out positive.

It is possible that when you plug in the upper limit you'll get a negative value but after subtracting the lower limit the result is positive.

 

[khadija112]

I have a few comments on your work/comments.

1) Did you draw the graph and see that the graph is above the x-axis between x=1 and x=2? If not, then how do you know that the area/answer should be positive? Or are all integrals supposed to be positive.

2) Yes, you are correct that if you interchange the limits of integration you must factor out a -1. My question is why would you want to do that given that in your case it makes the evaluation so awkward? Sure, if you already have a negative number factored out in front of the integral , then interchanging the limits will cause you do factor out a -1 making the number factored out positive.

It is possible that when you plug in the upper limit you'll get a negative value but after subtracting the lower limit the result is positive.

We were instructed to reverse the limits if the upper limit is an x value and the given value is less than the lower limit. So since I had to find g(1), and one is less than the lower limit of 2, I switched the limits.

 

[Steven G]

Since you get the same answer whether you switch the limits, you should switch the limits when it is convenient for you. I would only switch the limits if and only if I had a negative constant in front of the integral.