Is My Calculation for the Integral from -1 to 2 Correct?

[cse63146]

[SOLVED]Integration Problem

Homework Statement

$$\int \sqrt{(6t)^2 + (10t)^2} = \int \sqrt{36t^2 100t^2} = \int \sqrt{136t^2} = \sqrt{136} \int \sqrt{t^2} = \sqrt{136} \int t$$

Homework Equations

The Attempt at a Solution

Have I made a mistake anywhere? because its from -1 to 2, so I keep getting $$1.5\sqrt{136}$$ but it says it's wrong. Any ideas?

 

[Dick]

This a lot like integrating sqrt(t^2) from -1 to 1. If you simplify that to t, integrate to t^2/2 and put in the limits, you get 0. That's wrong. sqrt(t^2)=|t| NOT t. It's best to do the positive and negative ranges separately.

 

[cse63146]

So something like this:

$$\sqrt{136} ( \int t \ dt + \int -t \ dt)$$

where the first integral is from -1 to 1 and the second one is 1 to 2?

 

[sutupidmath]

-1 to 0 and from 0 to 2, since |t|=t, if t>0, and |t|=-t, if t<0

 

[cse63146]

Got it, thank you.