Problem 14 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{1}\left(x^{2}+1\right) e^{-x} d x\)

Step-by-Step Solution

Answer

The result of the integral is \(3 - \frac{6}{e}\).

Step 1: Identify the Method

The integral involves a polynomial and an exponential function. This suggests that integration by parts is an effective method to solve this integral, where we let one part be differentiated and another integrated.

Step 2: Choose Functions For Integration By Parts

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). Let \( u = x^2 + 1 \) and therefore \( du = 2x \, dx \). Let \( dv = e^{-x} \, dx \), which implies \( v = \int e^{-x} \, dx = -e^{-x} \).

Step 3: Apply Integration By Parts Formula

Apply the integration by parts formula, substituting back into the integral:\[ \int (x^2+1)e^{-x} dx = -(x^2+1)e^{-x} + \int 2x e^{-x} dx. \]

Step 4: Solve the Remaining Integral

Again, use integration by parts on the integral \( \int 2x e^{-x} dx \). Let \( u = 2x \) so \( du = 2 dx \), and \( dv = e^{-x} dx \) so \( v = -e^{-x} \).

Step 5: Compute the Second Integration By Parts

Apply the formula again:\[ \int 2x e^{-x} dx = -2xe^{-x} + \int 2e^{-x} dx \].

Step 6: Evaluate the Integral \( \int 2e^{-x} dx \)

Integrate \( 2e^{-x} \):\[ \int 2e^{-x} dx = -2e^{-x}. \]

Step 7: Substitute Back Into the Equation

Substitute \( \int 2x e^{-x} dx \) back into Step 5:\[ \int 2x e^{-x} dx = -2x e^{-x} - 2e^{-x}. \]

Step 8: Substitute Back Into the Original Equation

Combine everything:\[ \int (x^2 + 1) e^{-x} dx = -(x^2 + 1)e^{-x} - 2x e^{-x} - 2e^{-x}. \]

Step 9: Evaluate the Definite Integral from 0 to 1

Use the fundamental theorem of calculus to evaluate from 0 to 1:\[ [-(x^2 + 1)e^{-x} - 2x e^{-x} - 2e^{-x}]_{0}^{1} = [-(1 + 1)e^{-1} - 2 \cdot 1 \cdot e^{-1} - 2e^{-1}] - [-(0^2+1)e^{0} - 0 - 2 \, e^{0}]. \]

Step 10: Simplify the Results

Simplify the expressions:For 1: \(-(2)e^{-1} - 2e^{-1} - 2e^{-1} = -6e^{-1} \). For 0: \(-e^{0} - 2 = -3 \). So the result is \(-6e^{-1} + 3\).

Step 11: Compute the Final Result

Combine the terms and simplify further:\[ -\frac{6}{e} + 3. \]

Step 12: Final Answer Simplified

Thus, the result simplifies to \(3 - \frac{6}{e}\).

Key Concepts

Definite Integrals

A definite integral is a type of integral that deals with computing the area under the curve of a function between two specified limits, known as the bounds. In our problem, the integral is from 0 to 1. Definite integrals are powerful because they can accurately calculate the net "accumulated change" over an interval.

The fundamental theorem of calculus links definite integrals with antiderivatives. It states that if a function is continuous on \(a, b\), then the integral of the function from \(a\) to \(b\) is the difference between its antiderivative evaluated at these limits. For the integral \([a, b]\), we write it as \([F(b) - F(a)]\) where \(F\) is an antiderivative of the function being integrated.

In our exercise, we are calculating \([-(x^2+1)e^{-x} - 2xe^{-x} - 2e^{-x}]_{0}^{1}\). We substitute these bounds into the antiderivative expression to compute values separately, then subtract the value at the lower bound from the value at the upper bound. This process ensures we get the precise area or total value represented by the integral.

Exponential Functions

Exponential functions are functions where the variable is in the exponent position. They are commonly written in the form \(a^x\), but often involve the natural exponential function \(e^x\) in calculus. Exponential functions exhibit rapid growth or decay. They are highly relevant in sciences for modeling phenomena, such as population growth or radioactive decay.

In the context of our integration problem, \( ext{e}^{-x}\) represents an exponential decay. Integrating functions that include exponential components often requires special techniques like integration by parts, which hinges on simplification by rearrangement or decomposition of the original function. Here, integration by parts was needed because direct integration wasn't straightforward.

The derivative of \( ext{e}^{x}\) is \( ext{e}^{x}\), and correspondingly, the antiderivative or integral of \( ext{e}^{x}\) is still \( ext{e}^{x}\). Considering negative signs within exponentials, like in \( ext{e}^{-x}\), changes the nature of integration slightly but follows the same principles. Thus, \( ext{e}^{-x}\) integrates to \(- ext{e}^{-x}\). This feature plays a crucial role in deriving the antiderivatives during the integration process.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers of variables multiplied by coefficients. They are among the simplest and most flexible functions used in calculus. These functions typically take the form \(a_nx^n + a_{n-1}x^{n-1} + ... + a_0\), where \(a_n, a_{n-1}, ..., a_0\) are constants.

In our integral, \(x^2 + 1\) is a polynomial function. It's crucial to understand how polynomial functions behave during integration, as they serve as one of the building blocks in integration by parts. A polynomial is straightforward to differentiate and integrate, with each term \(ax^n\) having a derivative \(nax^{n-1}\) and an antiderivative \((a/(n+1))x^{n+1}\).

When integrating polynomials within the method of integration by parts, they often constitute the \(u\) part, for which the derivative \(du\) is straightforward to find. This assignment takes advantage of the polynomial's simplicity, allowing the more complex exponential component to be treated as \(dv\). In our exercise, decomposing the problem into these parts simplifies the integration process significantly, enabling us to integrate the combined polynomial and exponential function effectively.