$$\begin{array} {|c||c|c|c|c|} \hline x & 0 & 2 & 4 & 6 \\ \hline f(x) & 4 & k & 8 & 12 \\ \hline \end{array}$$

The function $f$ is continuous on the closed interval $[0,6]$ and has the values given in the table above. The trapezoidal approximation for $\int \limits_0^6 f(x)   dx$ found with 3 subintervals of equal length is 52. What is the value of $k$ ?

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It may be helpful to draw a quick sketch of the given points. The trapezoidal approximation is the sum of the areas of the three trapezoids shown below. The area of a trapezoid is the product of the average of the base lengths and the height.

$$A = \frac{1}{2}(b_1+b_2)h$$

Since the height $h$ of the trapezoids are 2 units each, the area of each trapezoid will just be the sum of the bases.

$$A = \frac{1}{2}(b_1+b_2)(2)$$ $$A = b_1+b_2$$

The lengths of the bases correspond to the $y$-values of the coordinates.

The sum of the areas need to be equal to 52 as stated in the problem.

$$4+k + k+8 +8+12 = 52$$ $$2k+32=52$$ $$2k = 20$$ $$k = \boxed{10}$$