To determine

a)

To write:

∫15(x+2x5)dx as limit of Riemann sum

Answer

5220

Explanation

1) Concept:

i) Definition of Riemann sum:

∫abfxdx=limn→∞⁡∑i=1nfxi* ∆x

Where,

∆x= b-an  and  xi=a+i ∆x

ii) The Fundamental Theorem of Calculus-Suppose f is continuous on [a, b]  then

∫abfxdx=Fb-F(a), F is an antiderivative of  f i.e F'=f

iii) Power rule of Integration:

∫xndx=xn+1n+1+C

2) Given:

∫15(x+2x5)dx

3) Calculation:

We have,

∫15(x+2x5)dx

Here a=1, b=5

∆x= b-an

= 5-1n

∆x=4n

xi=a+i ∆x

=1+i·4n

xi=1+4in

By Definition of Riemann sum,

∫15(x+2x5)dx=limn→∞⁡4n∑i=1n f1+4in

= limn→∞⁡4n∑i=1n1+4in+21+4in5

By using the command in Mathematica system,

=Limit[(4/n)*Sum[(1+(4*i/n))+2*(1+(4*i/n))^5,{i,1,n}],n→∞]

=5220

Conclusion:

Therefore,

∫15x+2x5dx=limn→∞⁡4n∑i=1n f1+4in=5220

To determine

b)

To check:

∫15(x+2x5)dx=5220

Answer

5220

Explanation

1) Calculation:

We have fx=x+2x5

Consider, F as an antiderivative of  f

Take antiderivative of  f,

Fx=x22+2·x66

Fx=x22+x63

By Fundamental Theorem of Calculus,

∫15x+2x5dx=F5-F1

=522+563-122+163

=252+156253-12-13

=242+156243

=12+5208

=5220

Conclusion:

Therefore,

∫15x+2x5dx=5220