How to solve this integral of an absolute function?

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

I think the answer for number 1 , graph somewhat like this

I get trouble for 2, 3, etc
I (k) = $\int_{-1}^{1} f(x) dx$
f(x) = $\mid x^2 - k^2 \mid$

2) k < 1
for negative side
$\int_{-1}^{-k} (x^2 - k^2) dx + \int_{-k}^{0} (k^2 - x^2) dx$
$\frac{x^3}{3} - k^2 {\mid} _{-1}^{-k} + (k^2 - \frac{x^3}{3} {\mid} )_{-k}^{0}$
$\frac{-k^3}{3} - k^2 - (\frac{-1}{3} - k^2) + k^2 - (k^2 - \frac{-k^3}{3} )$
$\frac{1}{3} - \frac{2k^3}{3}$

for positive side
$\int_{0}^{k} (k^2 - x^2) dx + \int_{k}^{1} ( x^2 - k^2 ) dx$
$k^2 - \frac{k^3}{3} - k^2 + \frac{1}{3} - k^2 - (\frac{k^3}{3} - k^2)$

$\frac{1}{3} - \frac{2k^2}{3}$

total = $- \frac{4k^3}{3} + \frac{2}{3}$​

but it is wrong.. why?

3) k > 1
for negative side

$\int_{-1}^{0} (k^2 - x^2) dx + \int_{0}^{1} (k^2 - x^2) dx$
but I get zero.
​

 

[DrClaude]

$$
\int_a^b (x^2-k^2) dx \neq \left[\frac{x^3}{3} - k^2 \right]_a^b
$$
The $k^2$ term is incorrect.

 

[Ray Vickson]

Homework Statement

View attachment 221277

Homework Equations

The Attempt at a Solution

I think the answer for number 1 , graph somewhat like this
View attachment 221278

I get trouble for 2, 3, etc
I (k) = $\int_{-1}^{1} f(x) dx$
f(x) = $\mid x^2 - k^2 \mid$

2) k < 1
for negative side
$\int_{-1}^{-k} (x^2 - k^2) dx + \int_{-k}^{0} (k^2 - x^2) dx$
$\frac{x^3}{3} - k^2 {\mid} _{-1}^{-k} + (k^2 - \frac{x^3}{3} {\mid} )_{-k}^{0}$
$\frac{-k^3}{3} - k^2 - (\frac{-1}{3} - k^2) + k^2 - (k^2 - \frac{-k^3}{3} )$
$\frac{1}{3} - \frac{2k^3}{3}$

for positive side
$\int_{0}^{k} (k^2 - x^2) dx + \int_{k}^{1} ( x^2 - k^2 ) dx$
$k^2 - \frac{k^3}{3} - k^2 + \frac{1}{3} - k^2 - (\frac{k^3}{3} - k^2)$

$\frac{1}{3} - \frac{2k^2}{3}$

total = $- \frac{4k^3}{3} + \frac{2}{3}$​

but it is wrong.. why?

3) k > 1
for negative side

$\int_{-1}^{0} (k^2 - x^2) dx + \int_{0}^{1} (k^2 - x^2) dx$
but I get zero.
​

You can simplify the analysis by noting that for any $k$ the function $f(x) = |x^2 - k^2|$ even (that is, $f(-x) = f(x)$), so $\int_{-1}^1 f(x) \, dx = 2 \int_0^1 f(x) \, dx$. You can see this from your graph, because you are trying to find the area under the graph from x = -1 to x = +1.

BTW: you are to be commended for now typing out your work---problem statements and solutions---instead of just posting images as you used to do. Why not go all the way and just dispense with the photos of the problem altogether? In this case they are so fuzzy as to be unreadable, and serve no purpose now anyway!