Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\ln 8} e^{x} d x$$

Short answer

Answer: The value of the definite integral \(\int_{0}^{\ln 8} e^x dx\) is 7.

Step-by-step solution

Find the antiderivative of the function

First, we need to find the antiderivative (indefinite integral) of the function \(e^x\). The derivative of the exponential function \(e^x\) is itself. Therefore, the antiderivative of \(e^x\) is also \(e^x\), let's denote it as F(x): $$F(x) = \int e^x dx = e^x + C$$

Evaluate the antiderivative at the limits of integration

Use the Fundamental Theorem of Calculus to evaluate the antiderivative at the limits of integration \(0\) and \(\ln 8\). According to the theorem: $$\int_{0}^{\ln 8} e^x dx = F(\ln 8) - F(0)$$ Now we need to evaluate the antiderivative at these points: $$F(\ln 8) = e^{\ln 8}$$ $$F(0) = e^0$$ Using the properties of exponentials and logarithms, we get: $$e^{\ln 8} = 8$$ $$e^0 = 1$$

Find the difference

Finally, find the difference between the antiderivative values at the limits of integration: $$\int_{0}^{\ln 8} e^x dx = F(\ln 8) - F(0) = 8 - 1 = 7$$ So, the value of the definite integral is: $$\int_{0}^{\ln 8} e^x dx = 7$$

Key concepts

Definite Integrals

Definite integrals are a core concept in calculus that represent the net area under a curve from one point to another. They're notable for their ability to quantify accumulation, such as distance traveled over a period of time when provided with a velocity function. Unlike indefinite integrals, which include a constant of integration (\( C \)), definite integrals yield a specific, calculable number.

To compute a definite integral, you typically find the antiderivative of the function, then apply the limits of integration—these are the points on the 'x' axis which you are measuring between. This process is made systematic by the Fundamental Theorem of Calculus which connects derivatives, antiderivatives, and definite integrals. For instance, in our original exercise \(\int_{0}^{\ln 8} e^{x} dx\), the definite integral is evaluated by finding the antiderivative of the function \(e^x\) and then computing the difference between the antiderivative values at the upper and lower limits, \(\ln 8\) and 0.

Antiderivatives

Antiderivatives are functions that reverse the process of differentiation, providing a function whose derivative is the original function. Finding an antiderivative is crucial for calculating definite integrals through the process highlighted by the Fundamental Theorem of Calculus. In general, to find the antiderivative \(F(x)\) of a function \(f(x)\), we seek a function such that \(F'(x) = f(x)\).

This theorem states that if \(F(x)\) is an antiderivative of \(f(x)\) over an interval, then the definite integral of \(f(x)\) from \(a\) to \(b\) is \(F(b) - F(a)\). In the given exercise, the function to integrate was the exponential function \(e^x\), whose antiderivative is itself. This characteristic simplifies the computation of definite integrals of exponential functions, as illustrated by the solution steps of the exercise.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In calculus, the exponential function \(e^x\) — where \(e\) is the mathematical constant approximately equal to 2.71828 — is especially important because it is its own derivative and antiderivative.

This unique property streamlines the process of integration as the antiderivative of \(e^x\) is simply \(e^x+C\), where \(C\) represents the constant of integration. In the context of our example, \(e^x\) is integrated over the interval from 0 to \(\ln 8\). Due to the properties of \(e\) and logarithms, \(e^{\ln 8}\) simplifies to 8 because raising \(e\) to the logarithm base \(e\) of any number returns that number. Similarly, \(e^0\) equals 1. These values are then used to find the final value of the definite integral, highlighting the interplay between antiderivatives and exponential functions in evaluating integrals.