# Thread: Trouble with a seemingly simple integral!

1. ## Trouble with a seemingly simple integral!

I'm doing a review sheet for my first exam coming up, and I thought this problem looked simple but I don't know how to get the correct answer (I have an answer key so I know I'm doing it wrong)

The definite integral from 0 to 6 of the absolute value of 4-x dx, I have to use limits to find it. Help me out here?

2. $\int_{0}^{6}{\left| 4-x \right|\,dx}=\int_{0}^{4}{\left| 4-x \right|\,dx}+\int_{4}^{6}{\left| 4-x \right|\,dx},$ thus $|4-x|=4-x$ whenever $0\le x\le4,$ and $|4-x|=x-4$ whenever $4\le x\le6$ and the integral is $\int_{0}^{6}{\left| 4-x \right|\,dx}=\int_{0}^{4}{(4-x)\,dx}+\int_{4}^{6}{(x-4)\,dx}=8+2=10.$

3. Thanks but I think I'm supposed to find it using a method we covered earlier , the one where you'd use limits to evaluate it. Could somebody perhaps outline how I'd do the problem that way?

4. Originally Posted by fattydq
Thanks but I think I'm supposed to find it using a method we covered earlier , the one where you'd use limits to evaluate it. Could somebody perhaps outline how I'd do the problem that way?
What you're asking here makes no sense. Krizalid has shown you how it needs to be done. Limits don't come into this at all (unless you want to evaluate it from first principles using Riemann sums).