Estimate the area under the graph of the function f(x) = 1/(x + 5) from x = 0 to x = 5 using a Riemann sun with n = 10 subintervals and left endpoints.

To estimate the area under the graph of the function \( f(x) = \frac{1}{x + 5} \) from \( x = 0 \) to \( x = 5 \) using a Riemann sum with \( n = 10 \) subintervals and left endpoints, we follow these steps:

1. **Determine the width of each subinterval (\(\Delta x\)):**

\[ \Delta x = \frac{b - a}{n} = \frac{5 - 0}{10} = 0.5 \]

2. **Identify the left endpoints:**

The left endpoints for each subinterval are:
\[ x_0 = 0, x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, x_4 = 2.0, x_5 = 2.5, x_6 = 3.0, x_7 = 3.5, x_8 = 4.0, x_9 = 4.5 \]

3. **Evaluate the function at each left endpoint:**

\[ f(x_i) = \frac{1}{x_i + 5} \]

So, we calculate:
\[ f(x_0) = \frac{1}{0 + 5} = \frac{1}{5} \]
\[ f(x_1) = \frac{1}{0.5 + 5} = \frac{1}{5.5} \]
\[ f(x_2) = \frac{1}{1.0 + 5} = \frac{1}{6} \]
\[ f(x_3) = \frac{1}{1.5 + 5} = \frac{1}{6.5} \]
\[ f(x_4) = \frac{1}{2.0 + 5} = \frac{1}{7} \]
\[ f(x_5) = \frac{1}{2.5 + 5} = \frac{1}{7.5} \]
\[ f(x_6) = \frac{1}{3.0 + 5} = \frac{1}{8} \]
\[ f(x_7) = \frac{1}{3.5 + 5} = \frac{1}{8.5} \]
\[ f(x_8) = \frac{1}{4.0 + 5} = \frac{1}{9} \]
\[ f(x_9) = \frac{1}{4.5 + 5} = \frac{1}{9.5} \]

4. **Multiply each function value by \(\Delta x\) and sum them up to get the Riemann sum:**

The Riemann sum \( L \) is:
\[ L = \sum_{i=0}^{9} f(x_i) \Delta x \]

\[ L = \left( \frac{1}{5} + \frac{1}{5.5} + \frac{1}{6} + \frac{1}{6.5} + \frac{1}{7} + \frac{1}{7.5} + \frac{1}{8} + \frac{1}{8.5} + \frac{1}{9} + \frac{1}{9.5} \right) \times 0.5 \]

Let's compute this:

\[ L \approx \left( 0.2 + 0.1818 + 0.1667 + 0.1538 + 0.1429 + 0.1333 + 0.125 + 0.1176 + 0.1111 + 0.1053 \right) \times 0.5 \]

\[ L \approx 1.4375 \times 0.5 \]

\[ L \approx 0.71875 \]

Thus, the estimated area under the graph of \( f(x) = \frac{1}{x + 5} \) from \( x = 0 \) to \( x = 5 \) using a Riemann sum with \( n = 10 \) subintervals and left endpoints is approximately \( 0.71875 \) square units.

Would you like more details or have any questions about this solution?

Here are 8 related questions to expand your understanding:

1. How would the estimation change if we used right endpoints instead of left endpoints?
2. What is the exact integral value of \( \int_{0}^{5} \frac{1}{x + 5} \, dx \)?
3. How does the number of subintervals (n) affect the accuracy of the Riemann sum?
4. What is a midpoint Riemann sum, and how would you calculate it for this function?
5. Can you derive the formula for the antiderivative of \( f(x) = \frac{1}{x + 5} \)?
6. How does the trapezoidal rule compare to the Riemann sum for this problem?
7. What are the differences between the left Riemann sum and the Simpson's rule for approximating integrals?
8. How would you estimate the area under the curve if the function were more complex, such as \( f(x) = \frac{1}{(x + 5)^2} \)?

**Tip:** When approximating areas under curves, increasing the number of subintervals generally improves the accuracy of your estimate.

Learn how to estimate the area under the graph of f(x) = 1/(x + 5) from x = 0 to x = 5 using a Riemann sum with 10 subintervals and left endpoints. Get a detailed step-by-step solution and understand the mathematical concepts involved. Suitable for college-level calculus students.

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