u-substitution? How to take this integral?

[sambellamy]

Hi, I am looking for the definite integral:

ʃ₀² ||t + t²|| dt

I have determined that the magnitude should be taken into consideration before calculating the integral - does that matter?

= ʃ₀² (t² + t⁴)^1/2 dt

should I use u - substitution? I know this is the opposite of the chain rule, but I am not sure where to go because the derivative of t² + t⁴ does not appear anywhere outside the parentheses.

Please help!

 

[Ishuda]

Hi, I am looking for the definite integral:

ʃ₀² ||t + t²|| dt

I have determined that the magnitude should be taken into consideration before calculating the integral - does that matter?

= ʃ₀² (t² + t⁴)^1/2 dt

should I use u - substitution? I know this is the opposite of the chain rule, but I am not sure where to go because the derivative of t² + t⁴ does not appear anywhere outside the parentheses.

Please help!

That particular integral is easy to solve, i.e.
\(\displaystyle 3 t \sqrt{1 + t^2}\)
is an exact differential. Unfortunately the norm of t + t² is not (t² + t⁴)^1/2

EDIT: Hint: Look at the limits of integration. What is the magnitude (absolute value) of the function in that interval.

 

[sambellamy]

I'm not sure what you mean by the norm? I was taking the magnitude of t + t² to be the square root of ( (t)² + (t²)² ), a la the Pythagorean theorem. What process did you use to arrive at [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT]?

I am not familiar with "exact differentials", and I think it might confuse the issue. How does one get to your solution without memorizing these?

Thanks

 

[Ishuda]

I'm not sure what you mean by the norm? I was taking the magnitude of t + t² to be the square root of ( (t)² + (t²)² ), a la the Pythagorean theorem. What process did you use to arrive at [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT]?

I am not familiar with "exact differentials", and I think it might confuse the issue. How does one get to your solution without memorizing these?

Thanks

The norm is, in this case, the same as the magnitude which is the same as the absolute value. In this case
||f(x)|| = [f²(x)]^1/2
or, for this case
||t + t²|| = [(t + t²)²]^1/2

As exact differential is a derivative of some function so that the integral of the derivative is the function plus an arbitrary constant, i.e.
\(\displaystyle 3 t \sqrt{1 + t^2} = \frac{d (1 + t^2)^{\frac{3}{2}}}{dt}\)
so
\(\displaystyle \int\, \sqrt{t^2 + t^4}\, dt = \frac{1}{3} \int 3\, t\, \sqrt{1 + t^2}\, dt = \frac{1}{3}\, \int \frac{d (1 + t^2)^{\frac{3}{2}}}{dt}\, \, dt = \frac{1}{3} (1 + t^2)^{\frac{3}{2}} + c\, \)

As I mentioned before, in a somewhat different way, what is the magnitude of t + t² in the interval [0,2]?

 

[sambellamy]

Aha! Thank you! I did not see that the t had simply been factored out before integration.

 

[HallsofIvy]

I'm not sure what you mean by the norm? I was taking the magnitude of t + t² to be the square root of ( (t)² + (t²)² ), a la the Pythagorean theorem.

This is a fundamental error! Yes \(\displaystyle |f(x)|= \sqrt{f(x)^2}\). But \(\displaystyle (t+ t^2)^2\) is NOT \(\displaystyle t^2+ t^4\). For any a, b, \(\displaystyle (a+ b)^2= a^2+ 2ab+ b^2\) so \(\displaystyle (t+ t^2)^2= t^2+ 2(t)(t^2)+ (t^2)^2= t^2+ 2t^3+ t^4\).

However, a more basic definition of the absolute value is "|x|= x if x is non-negative, -x if x is negative". Where is \(\displaystyle t+ t^2\) non-negative and where is it negative? Well, \(\displaystyle t+ t^2= t(t+ 1)\) and a product is positive if and only if both factors are positive- either t> 0 and t+ 1> 0 or t< 0 and t+ 1< 0. t+1> 0 is the same as t> -1. To be larger than both -1 and 0, t must be larger than 0. To be less than both 0 and -1, t must be less than -1.

\(\displaystyle t+ t^2\) is non-negative for t less than or equal to -1, negative for t between -1 and 0, and positive for t larger than 0. Since your integral is for t= 0 to 2, t is always larger than 0, \(\displaystyle t+ t^2\) is non-negative so \(\displaystyle |t+ t^2|= t+ t^2\). Your integral is the same as \(\displaystyle \int_0^2 t+ t^2 dt\).

\(\displaystyle \int_0^2 |t+ t^2| dt= \int_0^2 \sqrt{t^2+ 3t^3+ t^4}dt\)

Thanks

 

[sambellamy]

Hi HallsOfIvy, I understand that to square two added items is not to simply square both and add them together. The initial problem was a vector problem:

ʃ₀² ||ti + t²j|| dt

so that's where I got the magnitude thing. I had taken out the i and j and serendipitously, taking the integral as Ishuda showed allowed me to get the same answer as the book. I guess the vector context may or may not have been important, but it worked out in the end.