This is the situation:

square root trapezoidal

I have joined each of the points at the top of the vertical segments with a straight line.

Here, `a = 0` and `b = 1`, and the width of each trapezoid is given by:

`Delta x=frac{b-a}{n}=frac{1-0}{5}=0.2 `

`y_0= f(a)=` `f(0)=sqrt{0^2+1}=1`

`y_1= f(a + Δx)=` `f(0.2)=sqrt{0.2^2+1}=1.0198039`

`y_2= f(a + 2Δx)=` `f(0.4)=sqrt{0.4^2+1}=1.0770330`

`y_3= f(a + 3 Δx)=` `f(0.6)=sqrt{0.6^2+1}=1.1661904`

`y_4= f(a + 4 Δx)=` `f(0.8)=sqrt{0.8^2+1}=1.2806248`

`y_5= f(a + 5 Δx)=` `f(1.0)=sqrt{1^2+1}=1.4142136`

So we have:

`"Integral" ≈`

`0.2(1/2 times 1+1.0198039` `+1.0770330+1.1661904` `{:+1.2806248+1/2times 1.4142136)`

`=1.150`

So `int_0^1 sqrt{x^2+1}\ dx ≈ 1.150`

We can see in the graph above the trapezoids are very close to the original curve, so our approximation should be close to the real value. In fact, to 3 decimal places, the integral value is 1.148.

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