Chapter 6: Applications of Double Integration
Section 6.4: Average Value
Example 6.4.4
Find the average value of F=3 x2+2 y2+1 over R, the region bounded by the graphs of fx=arctanx+1−1/2 and gx=sinx on the interval x∈0,π/2.
See Example 6.2.4 and Example 6.1.4.
Solution
Mathematical Solution
The average value of F=3 x2+2 y2+1 over the region shown in Figure 6.1.4(a) is
∫0X∫sinxfxF ⅆy ⅆx+∫Xπ/2∫fxsinxF ⅆy ⅆx∫0X∫sinxfx1 ⅆy ⅆx+∫Xπ/2∫fxsinx1 ⅆy ⅆx
≐1.5724889590.3016523782
≐5.212917493
where X≐0.5059163080 is the x-coordinate of the intersection of y=fx with y=gx. The numerator is the volume computed in Example 6.2.4, while the denominator is the area computed in Example 6.1.4.
Maple Solution - Interactive
A solution from first principles entails simply formulating and evaluating the integrals for volume and area as found in Example 6.2.4 and Example 6.1.4, respectively.
Initialize
Context Panel: Assign Name
F=3 x2+2 y2+1→assign
f=arctanx+1−1/2→assign
g=sinx→assign
Obtain the intersections of the curves bounding R
Equate fx and gx, then press the Enter key.
Context Panel: Solve≻Numerically Solve
Context Panel: Assign to a Name≻X
f=g
arctan⁡x+1−12=sin⁡x
→solve
0.5059163080
→assign to a name
X
Obtain the average value of F
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
∫0X∫gfF ⅆy ⅆx+∫Xπ/2∫fgF ⅆy ⅆx∫0X∫gf1 ⅆy ⅆx+∫Xπ/2∫fg1 ⅆy ⅆx = 5.212917493
Maple Solution - Coded
Define F.
F≔3 x2+2 y2+1:
Define f.
f≔arctanx+1−1/2:
Define g.
g≔sinx:
Apply fsolve to f=g to determine X.
X≔fsolvef=g,x:
Use the Int and value commands.
Q≔IntF,y=g..f,x=0..X+IntF,y=f..g,x=X..π/2Int1,y=g..f,x=0..X+Int1,y=f..g,x=X..π/2
∫00.5059163080∫sin⁡xarctan⁡x+1−123⁢x2+2⁢y2+1ⅆyⅆx+∫0.505916308012⁢π∫arctan⁡x+1−12sin⁡x3⁢x2+2⁢y2+1ⅆyⅆx∫00.5059163080∫sin⁡xarctan⁡x+1−121ⅆyⅆx+∫0.505916308012⁢π∫arctan⁡x+1−12sin⁡x1ⅆyⅆx
q≔valueQ
5.212917493
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