In Problems 21–26, use the description of the region R to evaluate the indicated integral. 26. ∬ R x x 2 + y 2 d A ; R = { ( x , y ) | 0 ≤ x ≤ 4 y − y 2 , 0 ≤ y ≤ 2 }
In Problems 21–26, use the description of the region R to evaluate the indicated integral. 26. ∬ R x x 2 + y 2 d A ; R = { ( x , y ) | 0 ≤ x ≤ 4 y − y 2 , 0 ≤ y ≤ 2 }
Solution Summary: The author explains the value of the iterated integral 8sqrt2-63.
In Problems 21–26, use the description of the region R to evaluate the indicated integral.
26.
∬
R
x
x
2
+
y
2
d
A
;
R
=
{
(
x
,
y
)
|
0
≤
x
≤
4
y
−
y
2
,
0
≤
y
≤
2
}
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.
11. Find the area of the region between the function (*)=x -*+2 and the x-axis on the x-interval
[-1,2].
y= Vx-10
y = 0
12. Find the area between
and
between x =0 and x =9.
13. Find the area bounded by these two curves: Y =9-x?
2y = x² -3x +12
and
14. Find the area of the region bounded by these two curves: X = y" - 2y and *-y-4 =0
10. Find the area of the region bounded by the parabolas y- 6x = -x² and y + 2x = x² .
2)Let f(x) =2x-6
Let x0 be the x coordinate of x intercept of f(x) and y0 be the y coordinate of f(x) then
x0=
y0=
The area of the region bounded by f(x),x axis over interval [0,x0] is
Chapter 7 Solutions
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