Let f(x) = 42x²+ 25/6 , 0≤x≤6. The estimated area obtained by dividing the interval into 3 sub- intervals of equal length and the left end points of the sub-intervals for height of the rectangles is (in square
units)

To estimate the area under the curve \( f(x) = 42x^2 + \frac{25}{6} \) over the interval \([0, 6]\) using the left Riemann sum with 3 sub-intervals, follow these steps:

### Step 1: Determine the width of each sub-interval
Given that the interval \([0, 6]\) is divided into 3 sub-intervals of equal length:

\[
\text{Width of each sub-interval} = \Delta x = \frac{6 - 0}{3} = 2
\]

### Step 2: Identify the left endpoints
The left endpoints of the sub-intervals are:

- \( x_0 = 0 \)
- \( x_1 = 2 \)
- \( x_2 = 4 \)

### Step 3: Evaluate the function at the left endpoints
Calculate the function values at these points:

\[
f(x_0) = f(0) = 42(0)^2 + \frac{25}{6} = \frac{25}{6}
\]

\[
f(x_1) = f(2) = 42(2)^2 + \frac{25}{6} = 42(4) + \frac{25}{6} = 168 + \frac{25}{6} = \frac{1008 + 25}{6} = \frac{1033}{6}
\]

\[
f(x_2) = f(4) = 42(4)^2 + \frac{25}{6} = 42(16) + \frac{25}{6} = 672 + \frac{25}{6} = \frac{4032 + 25}{6} = \frac{4057}{6}
\]

### Step 4: Apply the left Riemann sum formula
The estimated area is calculated as:

\[
\text{Area} = \Delta x \cdot [f(x_0) + f(x_1) + f(x_2)]
\]

Substituting the values:

\[
\text{Area} = 2 \cdot \left[\frac{25}{6} + \frac{1033}{6} + \frac{4057}{6}\right] = 2 \cdot \frac{25 + 1033 + 4057}{6} = 2 \cdot \frac{5115}{6} = \frac{10230}{6} = 1705 \text{ square units}
\]

So, the estimated area is **1705 square units**.

Would you like more details or have any questions?

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Here are some related questions you might find useful:
1. How do you calculate a right Riemann sum?
2. What is the trapezoidal rule for estimating area under a curve?
3. How does increasing the number of sub-intervals affect the accuracy of Riemann sums?
4. What is the difference between left, right, and midpoint Riemann sums?
5. How do you derive the formula for the area under a parabola using integration?
6. Can Riemann sums be used to estimate volumes as well as areas?
7. How would you compute the definite integral of \( f(x) = 42x^2 + \frac{25}{6} \)?
8. What are the limitations of using Riemann sums for estimating areas?

**Tip:** The accuracy of a Riemann sum improves as the number of sub-intervals increases.

Learn how to estimate the area under the curve f(x) = 42x² + 25/6 over the interval [0, 6] using the left Riemann sum with 3 sub-intervals. This problem involves calculating Riemann sums, evaluating polynomial functions, and understanding area estimation techniques. Suitable for advanced high school students in Grades 11-12.

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