Definite Integral: Solving $y=\int^{10}_2 \frac{13.2}{x^{1.4}}$

[Karol]

Homework Statement

There is a problem in physics. i need to calculate the definite integral:
$$y=\int^{10}_2 \frac{13.2}{x^{1.4}}$$

Homework Equations

$$\int x^{-a}=\frac{1}{-a+1}x^{-a+1}$$

The Attempt at a Solution

$$y=\int^{10}_2 \frac{13.2}{x^{1.4}}=13.2\int^{10}_2 x^{-1.4}=13.2 \frac{1}{-0.4}x^{-0.4}|^{10}_2=-0.63$$
According to a graph i made with a graph software it came out -11.9, see picture

 

[Orodruin]

Neither of these answers can be correct, you have the integral of a positive quantity and the result must be positive. I suggest rechecking your arithmetics.

You really should also consider always writing out the dx in the integral for clarity. The expressions are mathematically incomplete without it.

 

[Karol]

$$y=\int^{10}_2 \frac{13.2}{x^{1.4}}dx=13.2\int^{10}_2 x^{-1.4}dx=$$
$$=13.2 \frac{1}{-0.4}x^{-0.4}|^{10}_2=-33\left(\frac{1}{10^{0.4}}-\frac{1}{2^{0.4}}\right)=11.9$$

 

[Orodruin]

That looks much better and is correct as far as I can tell.

 

[Karol]

Thanks

 

[Ray Vickson]

$$y=\int^{10}_2 \frac{13.2}{x^{1.4}}dx=13.2\int^{10}_2 x^{-1.4}dx=$$
$$=13.2 \frac{1}{-0.4}x^{-0.4}|^{10}_2=-33\left(\frac{1}{10^{0.4}}-\frac{1}{2^{0.4}}\right)=11.9$$

Please recognize that 11.9 is an approximation to the true answer (obtained by rounding to 3 significant figures); when you write "=11.9" you are hiding that fundamental fact, and are writing something that is not true. I think it is important that you broadcast your understanding of that issue by saying so explicitly---for example, by saying " ... = 11.9, rounded to 3 digits" or something similar. Even better would be to write "... ≈ 11.9 ..." or "... $\doteq$ 11.9 ...".

 

[Karol]

Right, i will use that notation next times, thanks