Express the definite integral as an infinite series. 7e dr

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**Task: Express the definite integral as an infinite series.**

\[ \int_0^1 7e^{-x^3} \, dx \]

This expression represents the definite integral of the function \(7e^{-x^3}\) from 0 to 1. To express this integral as an infinite series, one approach is to use the series expansion of the exponential function \(e^{-x^3}\).

### Steps to Express as Series:

1. **Expand \(e^{-x^3}\) as a Power Series:**

   The exponential function \(e^{-x^3}\) can be expanded using the Taylor series:

   \[ e^{-x^3} = \sum_{n=0}^{\infty} \frac{(-x^3)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{3n}}{n!} \]

2. **Multiply by the Constant:**

   Since the integrand is \(7e^{-x^3}\), multiply the series by 7:

   \[ 7 \cdot e^{-x^3} = \sum_{n=0}^{\infty} \frac{7(-1)^n x^{3n}}{n!} \]

3. **Integrate Term by Term:**

   Now, integrate term by term from 0 to 1:

   \[ \int_0^1 7e^{-x^3} \, dx = \sum_{n=0}^{\infty} \int_0^1 \frac{7(-1)^n x^{3n}}{n!} \, dx \]

   Evaluate each integral:

   \[ \int_0^1 x^{3n} \, dx = \left[\frac{x^{3n+1}}{3n+1}\right]_0^1 = \frac{1}{3n+1} \]

4. **Resulting Series:**

   The definite integral expressed as an infinite series is:

   \[ \sum_{n=0}^{\infty} \frac{7(-1)^n}{n!(3n+1)} \]

This expansion allows for the approximation of the integral to any desired accuracy by taking a sufficient number of terms from the series.
Transcribed Image Text:**Task: Express the definite integral as an infinite series.** \[ \int_0^1 7e^{-x^3} \, dx \] This expression represents the definite integral of the function \(7e^{-x^3}\) from 0 to 1. To express this integral as an infinite series, one approach is to use the series expansion of the exponential function \(e^{-x^3}\). ### Steps to Express as Series: 1. **Expand \(e^{-x^3}\) as a Power Series:** The exponential function \(e^{-x^3}\) can be expanded using the Taylor series: \[ e^{-x^3} = \sum_{n=0}^{\infty} \frac{(-x^3)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{3n}}{n!} \] 2. **Multiply by the Constant:** Since the integrand is \(7e^{-x^3}\), multiply the series by 7: \[ 7 \cdot e^{-x^3} = \sum_{n=0}^{\infty} \frac{7(-1)^n x^{3n}}{n!} \] 3. **Integrate Term by Term:** Now, integrate term by term from 0 to 1: \[ \int_0^1 7e^{-x^3} \, dx = \sum_{n=0}^{\infty} \int_0^1 \frac{7(-1)^n x^{3n}}{n!} \, dx \] Evaluate each integral: \[ \int_0^1 x^{3n} \, dx = \left[\frac{x^{3n+1}}{3n+1}\right]_0^1 = \frac{1}{3n+1} \] 4. **Resulting Series:** The definite integral expressed as an infinite series is: \[ \sum_{n=0}^{\infty} \frac{7(-1)^n}{n!(3n+1)} \] This expansion allows for the approximation of the integral to any desired accuracy by taking a sufficient number of terms from the series.
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