Problem 30 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(29-50.\) $$\int\left(2 e^{x}-3 e^{-2 x}\right) d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(2e^x - \frac{3}{2}e^{-2x} + C\).

Step 1: Break Down the Integral

The given integral is \(\int (2e^x - 3e^{-2x}) \, dx\). We can split this integral into two separate integrals: \(\int 2e^x \, dx - \int 3e^{-2x} \, dx\). This allows us to evaluate each part individually.

Step 2: Evaluate the First Integral

Start with the integral \(\int 2e^x \, dx\). Since \(e^x\) is its own derivative, the antiderivative of \(e^x\) is \(e^x\). Therefore, the antiderivative of \(2e^x\) is \(2e^x\).

Step 3: Evaluate the Second Integral

Now, evaluate \(\int 3e^{-2x} \, dx\). To find the antiderivative, note that the derivative of \(e^{-2x}\) is \(-2e^{-2x}\). To counteract the \(-2\), we multiply by \(-1/2\). Hence, the antiderivative of \(3e^{-2x}\) is \(-\frac{3}{2}e^{-2x}\).

Step 4: Combine the Results

Combine the results from Steps 2 and 3. The antiderivative of the original integral \(2e^x - 3e^{-2x}\) is \(2e^x - \frac{3}{2}e^{-2x} + C\), where \(C\) is the constant of integration.

Key Concepts

Antiderivative

The antiderivative, sometimes called the integral or primitive, is essentially the opposite of differentiating a function. If you take the derivative of a function and get back to where you started, you're dealing with an antiderivative. In mathematical terms, if you have a function \(f(x)\) and you know its derivative is \(f'(x)\), then the antiderivative of \(f'(x)\) is \(f(x) + C\), where \(C\) is a constant. While derivatives provide us with the slope of a function, antiderivatives help find the area under the curve or generally reverse the process of differentiation.

• For example, the antiderivative of \(e^x\) is \(e^x + C\) because the derivative of \(e^x\) is still \(e^x\).
• Similarly, since the derivative of \(e^{-2x}\) involves a factor of \(-2\), calculating its antiderivative requires reversing this factor.

Definite Integral

The definite integral represents the net area under a curve between two specific points on the x-axis. Unlike the antiderivative, which is more general, a definite integral gives a numerical value. This value tells us about total accumulation, such as distances or quantities over an interval.Given a function \(f(x)\), the definite integral from \(a\) to \(b\) is denoted as \[\int_{a}^{b} f(x) \, dx.\]This results in a specific number that represents the total "area" under \(f(x)\) from point \(a\) to point \(b\).Remember, while evaluating definite integrals, you don’t need a constant of integration (\(C\)) because it cancels out when you subtract the values of the antiderivative at the endpoints \(b\) and \(a\). Thus, definite integrals don't have a constant, and they focus on the total change or total area.

Indefinite Integral

An indefinite integral, in contrast to the definite integral, does not evaluate over a specific interval; instead, it generalizes the concept of antiderivatives. When you see an indefinite integral, \[\int f(x) \, dx,\]it is essentially asking you to find the antiderivative of the function \(f(x)\).Indefinite integrals result in a function rather than a single number. They contain a constant called the "constant of integration."

• The integral of each part, like in our exercise \(\int 2e^x - 3e^{-2x} \, dx\), is evaluated separately, allowing us to find their respective antiderivatives.
• This process helps us recompose a function from its derivative.

Indefinite integrals are used extensively in calculus to determine family functions of curves.

Constants of Integration

When dealing with indefinite integrals, you will always notice a "\(+ C\)" at the end of your expression. This is known as the constant of integration.This constant is crucial because it represents all possible vertical shifts of your antiderivative function on a graph. When you differentiate, all constants vanish (since the derivative of a constant is zero), leaving you with the same derivative no matter what constant you started with.

• For each indefinite integral, like our result \(2e^x - \frac{3}{2}e^{-2x} + C\), the \(+ C\) captures any constant-based information lost during differentiation.
• For physical, real-world applications, \(C\) can often represent initial values or conditions that were unknown or not needed for calculation purposes.

In sum, while it might seem minor, the constant of integration ensures that all solutions to an integral equation are accounted for.