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

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 \(k\) ?

Summary Submit Skip Question

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.

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