#### 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*