Integral of Absolute Value Function

[weiseguy]

Homework Statement

\int^{8}_{0}\left|x^{2} - 6x + 3\right|dx

This is for a single variable AP Calculus AB class in which we are solving using substitution method.

2. The attempt at a solution

I attempted it by just ignoring the abs value bars thinking that anything I am finding out "how much" of will be positive, and the negatives won't count. Solving with this in mind:

I came to the answer of 128/3

Please help as you can

 

[yaho8888]

when solving abs equations, first find the zeros and find out which part is negative, then do the integral for the individual part, when you get negative just take the abs value then add all the parts in the interval up.

 

[weiseguy]

So (I haven't checked), if the interval 0 to 3 is negative and 3 to 8 is positive. I would say the answer is the negative interval plus the positive one?

I need a little further explanation I think. We just started this in class and I had to miss half of the lecture.

 

[mutton]

What do you mean "negative interval"?

You know how the integral is the area above the curve minus the area below? Graph it and see.

 

[HallsofIvy]

Homework Statement

\int^{8}_{0}\left|x^{2} - 6x + 3\right|dx

This is for a single variable AP Calculus AB class in which we are solving using substitution method.

2. The attempt at a solution

I attempted it by just ignoring the abs value bars thinking that anything I am finding out "how much" of will be positive, and the negatives won't count. Solving with this in mind:

I came to the answer of 128/3

Please help as you can

Why would "negatives" not count? You can't ignore the abs value- you need to determine where x² - 6x+ 3 is positive and where it is negative. Start by determining where x²- 6x+ 3= 0.

 

[HallsofIvy]

So (I haven't checked), if the interval 0 to 3 is negative and 3 to 8 is positive. I would say the answer is the negative interval plus the positive one?

I need a little further explanation I think. We just started this in class and I had to miss half of the lecture.

You should have learned to solve quadratic equations long before a calculus class! I doubt that the part you missed was devoted to solving x²- 6x+ 3= 0! No, x= 3 is NOT a solution.

 

[weiseguy]

You should have learned to solve quadratic equations long before a calculus class! I doubt that the part you missed was devoted to solving x²- 6x+ 3= 0! No, x= 3 is NOT a solution.

I believe I stated "if for example the interval 0 to 3 was negative" not that it WAS indeed negative. It was 11pm and I didn't feel like doing any work at the moment. Basically though, I split the two up and add them together to find the total. It's like a displacement problem?