Math Problem Statement

Use a series to estimate the following​ integral's value with an error of magnitude less than

10 Superscript negative 510−5.

Integral from 0 to 0.4 sine x squared dx∫00.4sinx2 dx

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Part 1

Integral from 0 to 0.4 sine x squared dx∫00.4sinx2 dxalmost equals≈enter your response here

​(Do not round until the final answer. Then round to five decimal places as​ needed.)

Solution

To estimate the integral 00.4sin(x2)dx\int_0^{0.4} \sin(x^2) \, dx using a series with an error magnitude less than 10510^{-5}, let's proceed with the Taylor series expansion of sin(x2)\sin(x^2) around x=0x = 0:

  1. The Taylor series for sin(x)\sin(x) is: sin(x)=xx33!+x55!\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots

  2. By substituting x2x^2 for xx, we get the Taylor series for sin(x2)\sin(x^2): sin(x2)=x2x66+x10120\sin(x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \cdots

  3. To estimate 00.4sin(x2)dx\int_0^{0.4} \sin(x^2) \, dx, we integrate each term in the series expansion term by term: 00.4sin(x2)dx=00.4(x2x66+x10120)dx\int_0^{0.4} \sin(x^2) \, dx = \int_0^{0.4} \left( x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \cdots \right) \, dx

  4. Now, we integrate each term: 00.4x2dx=[x33]00.4=(0.4)33=0.0643=0.021333\int_0^{0.4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{0.4} = \frac{(0.4)^3}{3} = \frac{0.064}{3} = 0.021333 00.4x66dx=16[x77]00.4=16(0.4)77=0.001638467=3.91×105\int_0^{0.4} \frac{x^6}{6} \, dx = \frac{1}{6} \cdot \left[ \frac{x^7}{7} \right]_0^{0.4} = \frac{1}{6} \cdot \frac{(0.4)^7}{7} = \frac{0.0016384}{6 \cdot 7} = 3.91 \times 10^{-5}

Since 3.91×105<1053.91 \times 10^{-5} < 10^{-5}, we do not need additional terms to achieve the required accuracy.

  1. Adding the results: 00.4sin(x2)dx0.0213330.0000391=0.0212939\int_0^{0.4} \sin(x^2) \, dx \approx 0.021333 - 0.0000391 = 0.0212939

  2. Rounding to five decimal places: 00.4sin(x2)dx0.02129\int_0^{0.4} \sin(x^2) \, dx \approx 0.02129

Thus, the estimated value of the integral to five decimal places is: 0.02129\approx 0.02129

Would you like further clarification or more details on the series expansion?

Here are some follow-up questions to consider:

  1. How would the estimate change if we required an error of 10610^{-6} instead of 10510^{-5}?
  2. Could we use a different series expansion method to improve convergence?
  3. What are the general rules for determining the number of terms needed in a Taylor series approximation?
  4. How would the answer change if the interval were from 00 to 0.50.5 instead of 00 to 0.40.4?
  5. Can we apply Simpson's Rule to estimate this integral with the same accuracy?

Tip: Using the Taylor series method is effective for small intervals where the series converges quickly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Taylor Series Expansion
Integration
Error Bound in Series

Formulas

Taylor series for sin(x): sin(x) = x - x^3/3! + x^5/5! - ...
Taylor series for sin(x^2): sin(x^2) = x^2 - x^6/6 + x^10/120 - ...
Integral approximation using series expansion: ∫sin(x^2) dx ≈ ∫(x^2 - x^6/6 + x^10/120 - ...) dx

Theorems

Taylor Series Theorem
Error Bound in Taylor Series

Suitable Grade Level

College Calculus or Advanced High School (AP Calculus)