
To find:Approximating the area of region bounded by the graph of f and the x-axis over the specified interval using the indicated numbers n of rectangles of equal width and the exact area as n→∞ .

Answer to Problem 31E
Approximating the area of region with the number n of rectangles is
n | 4 | 8 | 20 | 50 | 100 | ∞ |
Approximate area | 40 | 38 | 36.8 | 36.32 | 36.16 | 36 |
Explanation of Solution
Given information:
The givenfunction is,
f(x)=2x+5
Given intervals [0,4]
And, rectangles are n
Concept used:
The area of a region;
Let f becontinuousandnonnegativeontheinterval [a,b] Thearea Aoftheregionboundedbythegraphof f ,the x-axis ,andtheverticallines x=a and x=b isgivenby
A=limx→∞n∑i=1f(a+(b−a)in)︸height(b−an)︸width
The givenfunctionis,
f(x)=2x+5
Thearea of the region bounded by the graph of f(x)=2x+5 and the x-axis between x=0 and x=4
Thedimensionsoftherectangles;
width ; b−an=4−0n=4n
And,
Height ; f(a+(b−a)in)=f(0+(4−0)in)=f(4in)
Given function; f(x)=2x+5
f(x)=2(4in)+5=8in+5=8i+5nn
Approximate the areas as the sum of the areas of n rectangles.
A=n∑i=1f(a+(b-a)in)(b-an)=n∑i=1(8i+5nn)(4n)=n∑i=132in2+20n
Applying Summationformulasandproperties.
n∑i=1i=n(n+1)2
Now,
A=32n2n∑i=1(i)+1nn∑i=120=32n2×n(n+1)2+20nn=16(n+1)n+20=36n+16n
Obtain a more accurate approximation of the area of the region by increasing the number of rectangles.
A(n)=36n+16n
Take,
n=4
A=36×4+164=1604=40
In this way, n=8,20,50,100 ....∞
The exact area by taking the limit as n approaches ∞
A=limx→∞[36n+16n]=36
Therefore, the exact area is 36 square units
Now, values of A=38, 36.8, 36.32,36.16 ....36
Finally, toapproximating the area of region by using values inthe following table.
n | 4 | 8 | 20 | 50 | 100 | ∞ |
Approximate area | 40 | 38 | 36.8 | 36.32 | 36.16 | 36 |
Chapter 12 Solutions
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