Question

Evaluate the definite integral. $$ \int_{0}^{1} e^{-2 x} d x $$

Short answer

The result of the definite integral \(\int_{0}^{1} e^{-2 x} d x\) is \(-\frac{1}{2e^2} + \frac{1}{2}\).

Step-by-step solution

Apply the Integration Rule

The first step is to apply the integral rule for \(e^{-2x}\). This is achieved by recognizing that the integral can be expressed as \(-\frac{1}{2} \cdot e^{-2x}\), since the derivative of \(e^{-2x}\) is indeed \(-2 e^{-2x}\). Thus, integrating the function over the interval [0,1] would result in evaluating \(-\frac{1}{2} \cdot e^{-2x}\) at the limits 0 and 1.

Evaluate at the Upper Limit

Plug in the upper limit 1 into the function to get \(-\frac{1}{2}e^{-2(1)}\), which simplifies to \(-\frac{1}{2e^2}\).

Evaluate at the Lower Limit

Now, plug in the lower limit of 0 into the function to get \(-\frac{1}{2}e^{-2(0)}\), which simplifies to \(-\frac{1}{2}\).

Subtract the Lower Limit from the Upper Limit

We find the difference between the two earlier computed limits by subtracting the result for the lower limit from the upper limit result. This results in \(-\frac{1}{2e^2} - -\frac{1}{2} = -\frac{1}{2e^2} + \frac{1}{2}\). This is in accordance with the fundamental theorem of calculus.

Key concepts

Exponential Functions

Exponential functions are everywhere in mathematics. They look like this: \(e^{x}\), where \(e\) is a mathematical constant approximately equal to 2.71828. These functions grow very fast and appear in many natural phenomena, like population growth or radioactive decay. In the exercise, we deal with a specific type of exponential function: \(e^{-2x}\). Here, the negative exponent \(-2x\) means the function decreases as \(x\) increases.

Exponential functions have a special property when it comes to calculus. They are the only functions whose derivative is proportional to the function itself. This makes them relatively easy to integrate. That's why \(e^{-2x}\) integrates to \(-\frac{1}{2}e^{-2x}\). The \(-\frac{1}{2}\) factor comes from reversing the effect of the chain rule, which is part of differentiation.

Integration by Substitution

Integration by substitution is a method that makes integrals easier to solve. Often, it's compared to the chain rule for derivatives, but in reverse. This method is perfect when you need to simplify a complex function or recognize part of the function's derivative.

Consider our integral \(\int e^{-2x} dx\). We can set \( u = -2x \), which means \( du = -2dx \) or \( dx = -\frac{1}{2}du \). By substituting, the original integral becomes \(-\frac{1}{2}\int e^{u} du\). This integral is straightforward—you simply integrate \( e^{u} \), which is itself, \( e^{u} + C \).

• Rather than memorizing rules, focus on recognizing when substitution can simplify your work.
• This method is particularly handy when dealing with exponential functions that have more complex exponents.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges two central concepts in calculus: differentiation and integration. It says that if you have a continuous function and you find its integral, then differentiating this integral gives you the original function back.

Cleverly using this theorem, we can evaluate definite integrals like \(\int_{0}^{1} e^{-2x} dx\). This involves finding an antiderivative (like we already did with \(-\frac{1}{2}e^{-2x}\)) and then calculating its value at the upper and lower limits.

• For our case, we evaluated \(-\frac{1}{2}e^{-2x}\) at \(x=1\) and \(x=0\).
• Finally, subtracting these results as per the theorem gives us the precise area under the curve \(e^{-2x}\) from \(x=0\) to \(x=1\).

Understanding this process helps demystify how integration can quantify areas under and between curves, making calculus a potent tool for analyzing mathematical and real-world problems.