Can you help me solving this trapezoidal rule

∫ [-1,1] (5x2 +1) dx

a) Using trapezoidal rule estimate the integral with n = 4 steps and find the upper bound for |ET|

T = ?
|ET| = ?

B) Evaluate the integral directly and find |ET| .
|ET| = ?

Could you plz show me the steps? Thanks.

a) f(x) = 5x² + 1

∆x = (1 − (−1)) / 4 = 1/2

x₀ = −1 -------> f(x₀) = 6
x₁ = −1/2 ----> f(x₁) = 9/4
x₂ = 0 --------> f(x₂) = 1
x₃ = 1/2 -----> f(x₃) = 9/4
x₄ = 1 -------> f(x₄) = 6

T = ∆x ∑[i=1 to 4] (f(xᵢ₋₁) + f(xᵢ))/2
T = ∆x/2 ∑[i=1 to 4] (f(xᵢ₋₁) + f(xᵢ))
T = 1/4 (f(x₀) + 2 f(x₁) + 2 f(x₂) + 2 f(x₃) + f(x₀))
T = 1/4 (6 + 9/2 + 2 + 9/2 + 6)
T = 23/4
T = 5.75

Suppose |f''(x)| ≤ K, then |ET| ≤ K (b−a)³/(12n²)

f'(x) = 10x
f''(x) = 10 ----> K = 10

|ET| ≤ 10 (1−(−1))³/(12*(4²)) = 10*8 / (12*16) = 5/12
|ET| ≤ 0.416666667

b)

∫₋₁¹ (5x² + 1) dx
= (5/3 x³ + x) |₋₁¹
= (5/3 + 1) − (−5/3 − 1)
= 16/3
= 5.333333333

|ET| = |16/3 − 23/4| = |−5/12| = 5/12 = 0.416666667

Let S = ∫ (5x^2 + 1) dx | -1 ≤ x ≤ 1

Using the trapezoidal rule,

Let f(x) = 5x^2 + 1

Δx = (b - a)/n

Δx = (1 + 1)/(4)

Δx = 1/2

Now, evaluate f( a + kΔx ) for every integer value of k from 0, until you have made n+1 steps:

f(-1) = 6

f(1/2 - 1) = f(-1/2) = 5/4 + 1 = 9/4

f(2/2 - 1) = f(0) = 1

f(3/2 - 1) = f(1/2) = 5/4 + 1 = 9/4

f(4/2 - 1) = f(1) = 6

According to the trapezoidal approximation:

S ≃ Δx [ f(a) + 2f(a + Δx) + 2f(a + 2Δx) + 2f(a + 3Δx) + ... + f(b) ]

S ≃ (1/2)[ 6 + 2(9/4) + 2(1) + 2(9/4) + 6 ]

S ≃ (1/2)[ 2(6) + 2(9/4) + 2(1) + 2(9/4) ]

S ≃ 6 + 9/4 + 1 + 9/4

S ≃ 7 + 9/2

S ≃ 23/2

The formula for error bounds is better left for looking up when you need it, since you are not expected to know how it's derived, and it shows up very seldomly in Calculus.

Bounds for Trapezoidal Error:

|ET| ≤ [ (b - a)^3 ]/[ 12n^2 ] * max ( f '' )

f ''(x) = 5

(So the max is just 5.)

|ET| ≤ 5[ (1 + 1)^3 ]/[ 12(4)^2 ]

|ET| ≤ 5[ 4 ]/[ 12(16) ]

|ET| ≤ 5/48

I'm just going to give you a link to notes on the Trapezoidal rule because I can't think of a way of doing the work on here that wouldn't be too hard to read. It is also just a matter of using a couple of formulas:
http://tutorial.math.lamar.edu/Classes/C…