Problem 54 - Calculus

Question

Evaluate the integrals. $$ \int_{0}^{1}\left(1-e^{x}\right)^{10} e^{x} d x $$

Step-by-Step Solution

Answer

\( \frac{(e-1)^{11}}{11} \)

Step 1: Recognize the Integral Type

Identify that the integral \( \int_{0}^{1} (1-e^{x})^{10} e^{x} \, dx \) is a definite integral with limits from 0 to 1.

Step 2: Substitution Method

Let us use substitution to simplify the integral. Set \( u = 1 - e^{x} \). Then \( du = -e^{x} \, dx \), so \( e^{x} \, dx = -du \).

Step 3: Change the Limits of Integration

With the substitution \( u = 1 - e^{x} \), change the limits of integration: when \( x = 0 \), \( u = 1 - e^0 = 0 \); when \( x = 1 \), \( u = 1 - e^1 = 1-e \).

Step 4: Rewrite the Integral

Using the substitution, the integral becomes:\[ -\int_{0}^{1-e} u^{10} \, du \]

Step 5: Evaluate the Integral

Integrate \( -\int_{0}^{1-e} u^{10} \, du \). The antiderivative of \( u^{10} \) is \( \frac{u^{11}}{11} \). Substituting limits,\[ -\left[ \frac{u^{11}}{11} \right]_{0}^{1-e} \]

Step 6: Calculate the Result

Compute \(-\left[ \frac{(1-e)^{11}}{11} - \frac{0^{11}}{11} \right]\ = \frac{(e-1)^{11}}{11} \).

Step 7: Simplify the Result

Since \((e-1)^{11}\) in absolute value is the symmetrical value to \((1-e)^{11}\) but positive, thus,\[ \frac{(e-1)^{11}}{11} \].

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify complex integrals. In this method, we transform the original variable into a new variable, making the integral easier to solve. For the given problem, we initially have \( \int (1-e^x)^{10} e^x \, dx \).

• We identify the substitution as \( u = 1 - e^x \). This is because the presence of \( (1-e^x)^{10} \) and \( e^x \) hints at a function and its derivative inside the integral.
• Calculating the derivative gives us \( du = -e^x \, dx \) or \( e^x \, dx = -du \).
• This substitution effectively transforms the integral from a complex form into a polynomial, \( \int u^{10} \, du \), which is simpler to integrate.

This technique not only simplifies the integration process but also helps in finding antiderivatives more easily in certain problems.

Limits of Integration

When using the substitution method, adjusting the limits of integration is crucial. The original integral has limits for \( x \) ranging from 0 to 1. However, with substitution \( u = 1 - e^x \), these limits must be converted:

• At \( x = 0 \), substitute to find \( u = 1 - e^0 = 0 \).
• At \( x = 1 \), substitute to find \( u = 1 - e^1 = 1-e \).

After substitution, the new integral is evaluated over the range \( u = 0 \) to \( u = 1-e \). Properly transforming limits is essential for accurate computation and ensures that when you switch back to the original variable, the values remain consistent.

Antiderivative

Finding the antiderivative is the core step in solving integrals. For our transformed integral \( \int u^{10} \, du \), we need to find its antiderivative. This involves:

• Recognizing the function \( u^{10} \) as a polynomial, which integrates straightforwardly using the formula \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \).
• Applying this to get the antiderivative, \( \frac{u^{11}}{11} \), which represents the indefinite integral of \( u^{10} \).

Remember, the antiderivative is a function that describes the accumulation of area beneath the curve of the original function \( u^{10} \). This is a critical step before we apply the definite integration limits.

Evaluate Integrals

After obtaining the antiderivative, the final challenge is evaluating the integral using the new limits. For the integral \( -\int_{0}^{1-e} u^{10} \, du \), this process involves:

• Inserting the upper limit \( u = 1-e \) into the antiderivative, giving \( \frac{(1-e)^{11}}{11} \).
• Inserting the lower limit \( u = 0 \), which results in \( \frac{0^{11}}{11} = 0 \).
• Calculating the difference, which in this case is \( -\left( \frac{(1-e)^{11}}{11} - 0 \right) \).

This calculation simplifies to \( \frac{(e-1)^{11}}{11} \). Evaluating integrals accurately relies on the proper application of limits to the antiderivative, concluding the process of integration and yielding the answer as per the problem's requirement.