Problem 17 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{1} \frac{y}{e^{2 y}} d y\)

Step-by-Step Solution

Answer

The evaluated integral is \( \frac{1 - 3e^{-2}}{4} \).

Step 1: Choose a Substitution

Since the integrand is a product of a polynomial \(y\) and an exponential function \(e^{2y}\), we use substitution to simplify the expression. Let \(u = 2y\), hence \(du = 2dy\). Therefore, \(dy = \frac{du}{2}\).

Step 2: Change the Limits of Integration

Since we have set \(u = 2y\), we should also change the limits of integration. When \(y = 0\), \(u = 2 \times 0 = 0\). When \(y = 1\), \(u = 2 \times 1 = 2\). The new limits with respect to \(u\) are from 0 to 2.

Step 3: Substitute and Simplify the Integral

Substitute into the integral using \(u = 2y\) and \(dy = \frac{du}{2}\). This transforms the integral into:\[ \int_{0}^{2} \frac{u/2}{e^u} \cdot \frac{du}{2} = \frac{1}{4} \int_{0}^{2} \frac{u}{e^u} du \]

Step 4: Use Integration by Parts

Apply integration by parts, where \(u = u\) and \(dv = \frac{1}{e^u} du\). Then, \(du = du\) and \(v = \int \frac{1}{e^u} du = -e^{-u}\). Integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Evaluate:\[\int \frac{u}{e^u} du = -u e^{-u} + \int e^{-u} du\]

Step 5: Simplify and Evaluate the Result

Evaluate \(\int e^{-u} du = -e^{-u}\). Substituting back:\[\int_{0}^{2} \frac{u}{e^u} du = -u e^{-u} - e^{-u} \big|_0^2\]Calculate:\[ = \left(-2e^{-2} - e^{-2}\right) - \left(0 + (-1)\right)\]\[= -3e^{-2} + 1\]

Step 6: Multiply by Constant Factor

Remember to multiply by the constant factor from substitution: \(\frac{1}{4}\). The final result is:\[\frac{1}{4} \left(-3e^{-2} + 1\right)\]Simplify to obtain the result:\[= \frac{-3e^{-2} + 1}{4}\]

Step 7: Final Answer

Thus, the evaluated integral is:\[ \frac{1 - 3e^{-2}}{4} \]

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus when dealing with products of functions. It's particularly useful when one of the functions becomes simpler when differentiated, and the other becomes manageable when integrated. The formula for integration by parts is an application of the product rule in reverse:\[\int u \, dv = uv - \int v \, du\]Here’s the breakdown:

• \(u\) is a function of the variable, and you differentiate \(u\) to get \(du\).
• \(dv\) is the rest of the integrand that you integrate to find \(v\).
• After finding \(v\), substitute back into the integration by parts formula.

For the example \(\int_{0}^{2} \frac{u}{e^u} du\), we applied this rule by choosing \(u = u\) and \(dv = \frac{1}{e^u} \ du\), which transformed the integral into a more straightforward expression. This simplification allows us to tackle integrals that are initially challenging to solve at first glance.

Definite Integral

A definite integral not only results in a number but also represents the net area under a curve from one point to another. In working with definite integrals, the limits of integration play a key role. They define the interval over which the function is integrated, providing an actual value rather than a function. In this exercise, the integral \(\int_{0}^{1} \frac{y}{e^{2y}} dy\) was evaluated by changing the limits of integration to accommodate the substitution, resulting in new limits from 0 to 2. This did not affect the integrand directly but was essential to compute the definite integral accurately.The main steps are:

• Change the integration bounds when substituting variables.
• Ensure all calculations reflect the updated bounds, yielding the correct numerical result.
• Carry out the final evaluation of the resulting expression at these endpoints.

Understanding and correctly applying the bounds ensures the definite integral is properly evaluated from start to finish, providing an accurate solution.

Integration Techniques

Integration techniques come in various forms, each tailored to simplify the task of finding antiderivatives. In calculus, these techniques provide different paths to evaluate integrals that can be difficult to solve with basic methods.Some common integration techniques include:

• Substitution: Replaces variables to simplify the integral, making it easier to evaluate. In our example, setting \(u = 2y\) transformed the original integral into a more manageable form.
• Integration by Parts: Often used when the integrand is a product of two functions. As demonstrated in this solution, it broke the integral into simpler pieces for easier integration.
• Partial Fraction Decomposition: Useful for integrating rational functions by breaking them into simpler fractions. Although not used here, it’s a powerful tool for other types of integrals.

Choosing the right integration technique depends on the specifics of the problem at hand. Combining these methods, as shown in the original exercise, can lead to successful evaluation even for seemingly complex integrals. Understanding when and how to apply these techniques is essential for any calculus student.