- #1
karush
Gold Member
MHB
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- 5
$\displaystyle \int_{-1}^{1} \int_{-2}^{3}(1-|x|) \,dy\,dx$
ok i was ? about the abs
ok i was ? about the abs
MarkFL said:The integrand is an even function, and the $x$ limits are symmetric, so apply the even function rule to remove the absolute value...:)
karush said:but some of this is in $-1 \le x \le 1$ how can you remove the abs
MarkFL said:\(\displaystyle I=2\int_{0}^{1}\left(1-x\right)\int_{-2}^{3}\,dy\,dx\)
Now iterate...:D
karush said:so then
$I = 2\left(\frac{1}{2}\right)(5) = 5$ ??
A double integral with absolute value in the integrand is a type of integral where the function being integrated includes the absolute value of a variable. This means that the area under the curve will be calculated differently depending on whether the variable is positive or negative.
The use of absolute value in a double integral allows for the calculation of the total area under a curve, regardless of whether the function is positive or negative. This is useful in cases where the function may change signs, as it ensures that the entire area is accounted for.
To calculate a double integral with absolute value in the integrand, the integral is split into two separate integrals: one for the positive values of the variable and one for the negative values. The two integrals are then added together to find the total area under the curve.
Double integrals with absolute value in the integrand are commonly used in physics, specifically in calculating the work done by a force over a distance. They are also used in economics, for finding the total cost or profit of a business, and in statistics, for calculating the total probability of an event.
One limitation of using double integrals with absolute value in the integrand is that they can be more complex and time-consuming to calculate compared to regular double integrals. They also require a deep understanding of calculus and may not be suitable for all types of functions.