Given information:

The given expression is ∫011+x4dx

Formula used:

Assume that f(x) is continuous on [a, b].

Let n be a positive Integer and △x=b−an.

If [a, b] is divided into n sub intervals, each of length △x with endpoint at P={x0,x1,x2,...,xn}.

  Tn=△x2(f(x0)+2f(x1)+2f(x2)+2f(xn−1)+f(xn)).then, limn→∞ Tn=∫abf(x)dx

Calculation:

Each subinterval has length

  △x=1−010△x=110△x=0.1

x0	0
x1	0.1
x2	0.2
x3	0.3
x4	0.4
x5	0.5
x6	0.6
x7	0.7
x9	0.8
x9	0.9
x10	1

The integral can be obtained as,

  ∫011+x4dx=0.12(f(0)+f(0.1)+2f(0.2)+2f(0.3)+2f(0.4)+2f(0.5)+2f(0.6)+2f(0.7)+2f(0.8)+2f(0.9)+f(1))∫011+x4dx=0.05×(1+1.000045+2.001598+2.008082+2.025438+2.061552+2.25652+2.2271954+2.3745314+2.57379+1.41424)∫011+x4dx=0.05×(21.9433668)∫011+x4dx=1.04716834

Given information:

The given expression is ∫011+x4dx .

Formula used:

Assume that f(x) is continuous on [a, b].

Let n be a positive Integer and △x=b−an.

If [a, b] is divided into n sub intervals, each of length △x and mi is the midpoint of the ith sub interval, set

  Mn=∑i=1nf(mi)△x.then limn→∞ Mn=∫abf(x)dx

Calculation: Each subinterval has length

  △x=1−010△x=110△x=0.1

x0	0
x1	0.1
x2	0.2
x3	0.3
x4	0.4
x5	0.5
x6	0.6
x7	0.7
x9	0.8
x9	0.9
x10	1

The integral value of the given integral can be obtained as,

  ∫011+x4dx=f(0+0.12)(0.1)+f(0.1+0.22)(0.1)+f(0.2+0.32)(0.1)+f(0.3+0.42)(0.1)+f(0.4+0.52)(0.1)+f(0.5+0.62)(0.1)+f(0.6+0.72)(0.1)+f(0.7+0.82)(0.1)+f(0.8+0.92)(0.1)+f(0.9+12)(0.1)∫011+x4dx=[f(0.05)(0.1)+f(0.15)(0.1)+f(0.25)(0.1)+f(0.35)(0.1)+f(0.45)(0.1)+f(0.55)(0.1)+f(0.65)(0.1)+f(0.75)(0.1)+f(0.85)(0.1)+f(0.95)(0.1)]∫011+x4dx=[0.1000005+0.100253+0.10019512+0.1007475+0.102029+0.104475+0.1085590+0.114734+0.123369+0.1347036]∫011+x4dx=1.0890702

Given information:

The given expression is ∫011+x4dx .

Formula used:

Assume that f(x) is continuous on [a, b].

Let n be a positive Integer and △x=b−an.

If [a, b] is divided into n sub intervals, each of length △x with endpoint at P={x0,x1,x2,...,xn}.

  Sn=△x3(f(x0)+4{f(x1)+f(x3)+f(x5)+...+f(xn−1)}+2{f(x2)+f(x4)+...+f(xn−2)}+f(xn)).then limn→∞ Sn=∫abf(x)dx.

Calculation:

Each subinterval has length

  △x=1−010△x=110△x=0.1

x0	0
x1	0.1
x2	0.2
x3	0.3
x4	0.4
x5	0.5
x6	0.6
x7	0.7
x9	0.8
x9	0.9
x10	1

The integral can be obtained as,

  ∫011+x4dx=0.13⋅[f(0)+4{f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9)}+2{f(0.2)+f(0.4)f(0.6)+f(0.8)}+f(1)]∫011+x4dx=0.033333⋅[1+4(1.000049+1.004041+1.030776+1.113597+1.286895)+2(1.000799+1.038644+1.062826+1.187265)+1.414213]∫011+x4dx=0.033333⋅[1+21.741252+8.578648+1.414213]∫011+x4dx=0.033333⋅(32.734145)∫011+x4dx=1.0911381