To determine
a)
To express:
The area under the curve
Answer
=limn→∞2n6∑i=1ni5
Explanation
1) Concept:
The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles.
A=limn→∞Rn=limn→∞fx1∆x+fx2∆x+…+fxn∆x=limn→∞∑i=1nfxi∆x
The width of the interval a, b is (b-a), so the width of each strip is
∆x=b-an
where x0=a and xn=b . The right endpoints of the subintervals are xi=a+i∆x.
2) Given:
y=x5, 0≤x≤2
3) Calculation:
It is given that
y=x5
Here a=0, b=2
∆x=b-an
=2-0n
∆x=2n
Now find out xi=a+i∆x
x1=0+12n=2n
x2=0+22n=4n
x3=0+32n=6n
xi=0+i·2n=2in
Therefore, the area under the function f(x) is given by
A=limn→∞Rn=limn→∞fx1∆x+fx2∆x+…+fxn∆x=limn→∞∑i=1nfxi∆x
=limn→∞∑i=1nfxi∆x
We have fx=x5, and substitute ∆x=2n.
=limn→∞∑i=1nxi5×2n
Substitute xi=2in
=limn→∞∑i=1n2in5×2n
=limn→∞∑i=1n2n6×i5
=limn→∞2n6∑i=1ni5
This is the area under the given curve.
Conclusion:
A=limn→∞2n6∑i=1ni5
To determine
b)
To find:
The sum in expression of the area under the curve
Answer
n2n+122n2+2n-112
Explanation
1) Calculation:
From part a),
A=limn→∞2n6∑i=1ni5
Now need to simplify the summation part.
That means to simplify ∑i=1ni5
By using the Command in Mathematica system,
Sum[i^5,{i,1,n}],
∑i=1ni5=112n2(1+n)2(-1+2n+2n2)
It can be written as,
∑i=1ni5=n2n+122n2+2n-112
Conclusion:
∑i=1ni5=n2n+122n2+2n-112
To determine
c)
To evaluate:
limn→∞2n6∑i=1ni5
Answer
323
Explanation
1) Calculation:
From part a) and b),
A=limn→∞2n6∑i=1ni5
∑i=1ni5=n2n+122n2+2n-112
Substitute for the summation part, and evaluate the limit.
A=limn→∞2n6×n2n+122n2+2n-112
=limn→∞64n6×n2n+122n2+2n-112
=limn→∞6412×n+122n2+2n-1n4
=limn→∞6412×n2+2n+12n2+2n-1n4
=limn→∞163×n2+2n+1n2×2n2+2n-1n2
=limn→∞163n2n2+2nn2+1n22n2n2+2nn2-1n2
=163limn→∞1+2n+1n22+2n-1n2
Simplify the limit.
=1631+0+02+0-0
=1632
=323
Conclusion:
limn→∞2n6∑i=1ni5=323