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\int_{0}^{4} \sqrt{x^{2}+2} d x

Approximate the definite integral wit...

Feb 2, 2024

\int_{0}^{4} \sqrt{x^{2}+2} d x

Approximate the definite integral with the Trapezoid Rule and

\boldsymbol{n}=\mathbf{6}

Generated Graph

Solution by Steps

step 1

To approximate the integral

\int_{0}^{4} \sqrt{x^{2}+2} dx

using the trapezoidal rule with

n=6

, we first divide the interval

[0, 4]

into 6 equal parts

step 2

The width of each trapezoid (

h

) is

\frac{4-0}{6} = \frac{2}{3}

step 3

Calculate the values of the function at the endpoints and the midpoints:

f(x_i) = \sqrt{x_i^2 + 2}

for

x_i = 0, \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3}, \frac{10}{3}, 4

step 4

Apply the trapezoidal rule formula:

\text{Trapezoidal Sum} = \frac{h}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + 2f(x_4) + 2f(x_5) + f(x_6)]

step 5

Substitute the values of

h

and

f(x_i)

into the formula to get the approximate value of the integral

Answer

The approximate value of the integral using the trapezoidal rule with

n=6

is 10.283.

Key Concept

Trapezoidal Rule for Approximating Integrals

Explanation

The trapezoidal rule is a numerical method to approximate the value of a definite integral. It works by dividing the area under the curve into trapezoids rather than rectangles (as in the midpoint or left/right endpoint methods) and better approximates areas under curves that are non-linear.