Math Problem Statement

Estimate the area under the graph of the function f(x) = sqrt(x + 5) from x = -2 to x = 5 using a Riemann sun with n = 10 subintervals and left endpoints.

Solution

To estimate the area under the graph of the function f(x)=x+5f(x) = \sqrt{x + 5} from x=2x = -2 to x=5x = 5 using a Riemann sum with n=10n = 10 subintervals and left endpoints, follow these steps:

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

Δx=ban=5(2)10=710=0.7\Delta x = \frac{b - a}{n} = \frac{5 - (-2)}{10} = \frac{7}{10} = 0.7

  1. Identify the left endpoints:

The left endpoints for each subinterval are: x0=2,x1=1.3,x2=0.6,x3=0.1,x4=0.8,x5=1.5,x6=2.2,x7=2.9,x8=3.6,x9=4.3x_0 = -2, x_1 = -1.3, x_2 = -0.6, x_3 = 0.1, x_4 = 0.8, x_5 = 1.5, x_6 = 2.2, x_7 = 2.9, x_8 = 3.6, x_9 = 4.3

  1. Evaluate the function at each left endpoint:

f(xi)=xi+5f(x_i) = \sqrt{x_i + 5}

So, we calculate: f(x0)=2+5=3f(x_0) = \sqrt{-2 + 5} = \sqrt{3} f(x1)=1.3+5=3.7f(x_1) = \sqrt{-1.3 + 5} = \sqrt{3.7} f(x2)=0.6+5=4.4f(x_2) = \sqrt{-0.6 + 5} = \sqrt{4.4} f(x3)=0.1+5=5.1f(x_3) = \sqrt{0.1 + 5} = \sqrt{5.1} f(x4)=0.8+5=5.8f(x_4) = \sqrt{0.8 + 5} = \sqrt{5.8} f(x5)=1.5+5=6.5f(x_5) = \sqrt{1.5 + 5} = \sqrt{6.5} f(x6)=2.2+5=7.2f(x_6) = \sqrt{2.2 + 5} = \sqrt{7.2} f(x7)=2.9+5=7.9f(x_7) = \sqrt{2.9 + 5} = \sqrt{7.9} f(x8)=3.6+5=8.6f(x_8) = \sqrt{3.6 + 5} = \sqrt{8.6} f(x9)=4.3+5=9.3f(x_9) = \sqrt{4.3 + 5} = \sqrt{9.3}

  1. Multiply each function value by Δx\Delta x and sum them up to get the Riemann sum:

The Riemann sum LL is: L=i=09f(xi)ΔxL = \sum_{i=0}^{9} f(x_i) \Delta x

L=(3+3.7+4.4+5.1+5.8+6.5+7.2+7.9+8.6+9.3)×0.7L = \left( \sqrt{3} + \sqrt{3.7} + \sqrt{4.4} + \sqrt{5.1} + \sqrt{5.8} + \sqrt{6.5} + \sqrt{7.2} + \sqrt{7.9} + \sqrt{8.6} + \sqrt{9.3} \right) \times 0.7

Let's compute this:

L(1.732+1.923+2.097+2.258+2.408+2.550+2.683+2.810+2.932+3.050)×0.7L \approx \left( 1.732 + 1.923 + 2.097 + 2.258 + 2.408 + 2.550 + 2.683 + 2.810 + 2.932 + 3.050 \right) \times 0.7

L24.443×0.7L \approx 24.443 \times 0.7

L17.1101L \approx 17.1101

Thus, the estimated area under the graph of f(x)=x+5f(x) = \sqrt{x + 5} from x=2x = -2 to x=5x = 5 using a Riemann sum with n=10n = 10 subintervals and left endpoints is approximately 17.110117.1101 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 25x+5dx\int_{-2}^{5} \sqrt{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)=x+5f(x) = \sqrt{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)=(x+5)3/2f(x) = (x + 5)^{3/2}?

Tip: When calculating Riemann sums, using a larger number of subintervals (n) generally provides a more accurate estimate of the integral.

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Math Problem Analysis

Mathematical Concepts

Calculus
Riemann Sum

Formulas

Riemann sum formula
Area estimation under curve

Theorems

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Suitable Grade Level

College Level