Step-by-step explanation:
A left Riemann sum approximates a definite integral as:
[tex]\int\limits^b_a {f(x)} \, dx \approx \sum\limits_{k=1}^{n}f(x_{k}) \Delta x \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times (k-1)[/tex]
Given ∫₂⁸ cos(x²) dx:
a = 2, b = 8, and f(x) = cos(x²)
Therefore, Δx = 6/n and x = 2 + (6/n) (k − 1).
Plugging into the sum:
∑₁ⁿ cos((2 + (6/n) (k − 1))²) (6/n)
Therefore, the answer is C. Notice that answer D would be a right Riemann sum rather than a left (uses k instead of k−1).
