What is the approximation of $\int_0^6 h(x)\,dx$ when a left rectangle Riemann sum is used with 3 subintervals?

Key takeaway: This is a left Riemann sum problem. The key idea is to multiply each subinterval width by the function value at the left endpoint, then add the three rectangle areas.

Using a left rectangle Riemann sum with 3 subintervals, the approximation is the sum of the left-endpoint heights times each subinterval width.

From the given setup:

• first rectangle: $2\cdot 1$
• second rectangle: $2\cdot 1$
• third rectangle: $3\cdot 0$

So,

$\int_0^6 h(x)\,dx \approx (2\cdot 1)+(2\cdot 1)+(3\cdot 0)=4.$

Approximation: $4$

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Pitfalls the pros know 👇 A common mistake is using right endpoints instead of left endpoints. In a left Riemann sum, each rectangle height must come from the left side of each subinterval, not the right side or midpoint.

What if the problem changes? If the same table were used with a right Riemann sum, the approximation could change because the heights would come from the right endpoints. If the subintervals were not equal, each rectangle would still be computed as $\text{width} \times \text{left height}$, but the widths might differ.

Tags: left Riemann sum, subinterval, rectangle approximation