Evaluate the integral by making an appropriate change of variables. ∬ R y − 4 x y + 4 x d A , where R is the region enclosed by the lines y = 4 x , y = 4 x − 1 − 2 , y = 2 − 4 x , y = 5 − 4 x .
Evaluate the integral by making an appropriate change of variables. ∬ R y − 4 x y + 4 x d A , where R is the region enclosed by the lines y = 4 x , y = 4 x − 1 − 2 , y = 2 − 4 x , y = 5 − 4 x .
Evaluate the integral by making an appropriate change of variables.
∬
R
y
−
4
x
y
+
4
x
d
A
,
where R is the region enclosed by the lines
y
=
4
x
,
y
=
4
x
−
1
−
2
,
y
=
2
−
4
x
,
y
=
5
−
4
x
.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
Find // 4x y dA, where R is a region enclosed by 3x – 2y = 0,3x – 2y = 1, x + 3y = 0, and x + 3y = 6.
R
Use the change of variables u = 3x – 2y and v = x + 3y.
(Use symbolic notation and fractions where needed.)
4x² ydA =
R
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