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Chapter 7: Problem 22

Evaluate the following integrals or state that they diverge. $$\int_{-\infty}^{a} \sqrt{e^{x}} d x, a \text { real }$$

Short Answer

Expert verified
If it converges, find its value. Answer: The integral converges, and its value is $2e^{\frac{1}{2}a}$.

Step by step solution

01

Simplify the integrand

First, simplify the integrand \(\sqrt{e^x}\) using the property \(a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}\). So we have: $$\sqrt{e^x} = \left(e^{x}\right)^{\frac{1}{2}} = e^{\frac{1}{2}x}$$ Now, the integral becomes: $$\int_{-\infty}^{a} e^{\frac{1}{2}x} d x$$
02

Find the antiderivative

To find the antiderivative of \(e^{\frac{1}{2}x}\), we can use the general formula, which is: $$\int e^{kx} dx = \frac{1}{k} e^{kx} + C$$ where \(k\) is a constant, in our case \(k = \frac{1}{2}\): $$\int e^{\frac{1}{2}x} dx = 2e^{\frac{1}{2}x} + C$$
03

Evaluate the definite integral

Now, we can evaluate the definite integral using the antiderivative and the given limits of integration: $$\int_{-\infty}^{a} e^{\frac{1}{2}x} d x = \left[2e^{\frac{1}{2}x}\right]_{-\infty}^{a} = 2e^{\frac{1}{2}a} - \lim_{x \to -\infty} 2e^{\frac{1}{2}x}$$ As \(x \to -\infty\), the term \(e^{\frac{1}{2}x}\) approaches 0, so the limit converges: $$\lim_{x \to -\infty} 2e^{\frac{1}{2}x} = 0$$ Therefore, the definite integral converges, and we can find its value: $$\int_{-\infty}^{a} e^{\frac{1}{2}x} d x = 2e^{\frac{1}{2}a} - 0 = 2e^{\frac{1}{2}a}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative
The antiderivative is an essential concept in calculus, often referred to as the indefinite integral. It represents the reverse process of differentiation, finding a function whose derivative is the original function given.
In the context of the exercise, we identified the function under the integral sign as \(e^{\frac{1}{2}x}\). To find the antiderivative of an exponential function like this, we use a standard formula: \(\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C\). Here, \(k\) is a constant. Therefore, the antiderivative of \(e^{\frac{1}{2}x}\) is \(2e^{\frac{1}{2}x} + C\), where \(C\) is the integration constant.
The importance of finding an antiderivative lies in solving definite integrals, as it allows us to calculate the total area under the curve, or in practical situations, the total quantity accumulated.
Exponential Function
Exponential functions are one of the most common types of functions in mathematics. The general form is expressed as \(e^{kx}\), where \(e\) is Euler's number, approximately 2.71828, and \(k\) is a constant multiplier. This type of function shows rapid growth or decay.
In our exercise, the function \(\sqrt{e^x} = e^{\frac{1}{2}x}\) is an exponential function. The exponential nature is maintained but simplified through the property \(\sqrt{e^{x}} = e^{\frac{1}{2}x}\), which helps to efficiently compute integrals.
Understanding exponential functions is vital in numerous fields such as mathematics, science, and engineering because they describe phenomena related to growth and decay, such as population growth, radioactive decay, and compound interest.
Definite Integral
A definite integral computes the accumulated value of a function over a specified interval. It is often used to find areas under curves or the total value a function assumes between two points.
In this exercise, we focus on the improper definite integral \(\int_{-\infty}^{a} e^{\frac{1}{2}x} \, dx\), where the lower limit is negative infinity. Improper integrals take into account limits that approach infinity or points where the function is undefined.
To evaluate a definite integral, you first find the antiderivative, then apply the limits of integration. The difference of the antiderivative evaluated at the upper and lower limits gives the result of the integral. In this case, the lower limit contribution \(\lim_{x \to -\infty} 2e^{\frac{1}{2}x}\) converges to 0, resulting in a finite value of \(2e^{\frac{1}{2}a}\). Thus, the integral converges and gives meaningful results.

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