Problem 10 - Calculus for Biology and Medicine

Question

Evaluate the indefinite integral by making the given substitution. $$ \int 3 e^{1-x} d x, \text { with } u=1-x $$

Step-by-Step Solution

Answer

The indefinite integral evaluates to \( -3 e^{1-x} + C \).

Step 1: Identify the substitution

The problem suggests using the substitution \( u = 1 - x \). This means that we are changing the variable of integration from \( x \) to \( u \).

Step 2: Compute the differential

To substitute \( u \) for \( x \), we need to express \( dx \) in terms of \( du \). Differentiating \( u = 1 - x \), we get \( du = -dx \). Thus, \( dx = -du \).

Step 3: Rewrite the integral using substitution

Now substitute \( u = 1 - x \) and \( dx = -du \) into the integral: \[ \int 3 e^{1-x} dx = \int 3 e^u (-du) \]This simplifies to:\[ \int -3 e^u du \]

Step 4: Integrate the new integral

The integral \( \int -3 e^u du \) can be calculated as:\[ -3 \int e^u du = -3 e^u + C \]where \( C \) is the constant of integration.

Step 5: Substitute back in terms of x

Finally, substitute back \( u = 1 - x \) to express the antiderivative in terms of \( x \):\[ -3 e^{1-x} + C \]

Key Concepts

The Substitution Method in Integration

The substitution method is a powerful tool in calculus to simplify the process of integration. When faced with a complex integral, substitution allows us to transform the integrand into a simpler form. This is done by changing the variable of integration, often through a clever choice that simplifies the function.
For example:

• If you're given an integral with a composite function, substitution can simplify the process.
• By choosing the right substitution, you can often transform a difficult integral into a standard form that's easier to evaluate.

In our exercise, we begin by identifying the substitution: let \( u = 1 - x \). This changes our variable of integration from \( x \) to \( u \) and requires us to also find the differential \( du \).
Differentiating \( u \) with respect to \( x \), we get \( du = -dx \), or equivalently \( dx = -du \). With this substitution, the integral is simplified and more manageable.

Understanding Definite Integrals

While the exercise provided deals with an indefinite integral, it's helpful to understand definite integrals, which represent the computation of the integral over a specific range of values. A definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively.
If you were to solve a similar problem with definite limits, the substitution process still applies, but you'll need to change the limits accordingly:

• When you substitute \( x \) with \( u \), calculate the new limits in terms of \( u \).
• Evaluate the integral with the transformed limits using the new variable.

For instance, if our integral from the exercise had limits, say from \( x = 0 \) to \( x = 1 \), upon substituting \( u = 1 - x \), the limits would change. As a result, \( x = 0 \) becomes \( u = 1 \), and \( x = 1 \) becomes \( u = 0 \), flipping the limits, which adds a new layer of consideration to the evaluation process.

Differential Calculus and Integrals

Differential calculus and integral calculus are two sides of the same coin, where each process undoes the other to some extent. Understanding how differentiation relates to integration deepens comprehension.

• Differentiation deals with determining the instantaneous rate of change of a function, or the slope of its curve at any point.
• Integration, on the other hand, is often interpreted as finding the total accumulation or area under the curve of a function.

In practice, when you perform substitution in an integral, you're differentiating the chosen substitution equation to find \( du \). For our exercise, starting with \( u = 1 - x \), we found \( du = -dx \). This differential relationship is a direct application of the reverse process of differentiation.
When integrating \( -3 e^u \, du \), we find the antiderivative, which is the reverse of differentiation. This gives \( -3 e^u + C \), where reversing the substitution leads to the final solution \( -3 e^{1-x} + C \).