Using n=4 trapezoids, how do you approximate the value of #int sqrt(x+1) dx# from [1,3]?
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To approximate the value of ( \int_{1}^{3} \sqrt{x + 1} , dx ) using ( n = 4 ) trapezoids:

Divide the interval ([1, 3]) into ( n = 4 ) equal subintervals. Since we have 4 trapezoids, each subinterval width will be ( \Delta x = \frac{3  1}{4} = 0.5 ).

Calculate the function values at the endpoints of each subinterval. In this case, calculate ( \sqrt{x + 1} ) at ( x = 1, 1.5, 2, 2.5, 3 ).

Use the trapezoidal rule formula to find the area of each trapezoid: ( A_i = \frac{h}{2} [f(x_i) + f(x_{i+1})] ), where ( h ) is the width of each subinterval, ( f(x_i) ) is the function value at the left endpoint, and ( f(x_{i+1}) ) is the function value at the right endpoint.

Sum up the areas of all the trapezoids to get the approximate value of the integral: ( \int_{1}^{3} \sqrt{x + 1} , dx \approx A_1 + A_2 + A_3 + A_4 ).

Calculate each ( A_i ) using the formula from step 3.

Add up the values of ( A_i ) obtained in step 5 to get the final approximation of the integral.
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