232.q1.2c Double integral with absolute value in integrand

[karush]

$\displaystyle \int_{-1}^{1} \int_{-2}^{3}(1-|x|) \,dy\,dx$

ok i was ? about the abs

 

[MarkFL]

Re: 232.q1.2c dbl int with abs

The integrand is an even function, and the $x$ limits are symmetric, so apply the even function rule to remove the absolute value...:)

 

[karush]

Re: 232.q1.2c dbl int with abs

The integrand is an even function, and the $x$ limits are symmetric, so apply the even function rule to remove the absolute value...:)

but some of this is in $-1 \le x \le 1$ how can you remove the abs

 

[MarkFL]

Re: 232.q1.2c dbl int with abs

but some of this is in $-1 \le x \le 1$ how can you remove the abs

We are given:

\(\displaystyle I=\int_{-1}^{1}\int_{-2}^{3} 1-|x|\,dy\,dx\)

Now, since the integrand is a function of $x$ only, we can write:

\(\displaystyle I=\int_{-1}^{1}\left(1-|x|\right)\int_{-2}^{3}\,dy\,dx\)

Apply the even function rule:

\(\displaystyle I=2\int_{0}^{1}\left(1-x\right)\int_{-2}^{3}\,dy\,dx\)

Now iterate...:D

 

[karush]

Re: 232.q1.2c dbl int with abs

\(\displaystyle I=2\int_{0}^{1}\left(1-x\right)\int_{-2}^{3}\,dy\,dx\)

Now iterate...:D

so then

$I = 2\left(\frac{1}{2}\right)(5) = 5$ ??

 

[MarkFL]

Re: 232.q1.2c dbl int with abs

so then

$I = 2\left(\frac{1}{2}\right)(5) = 5$ ??

\(\displaystyle I=2\int_{0}^{1}\left(1-x\right)\int_{-2}^{3}\,dy\,dx=10\int_{0}^{1}\left(1-x\right)\,dx=5\left[2x-x^2\right]_0^1=5(2-1)=5\quad\checkmark\)