/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 41

Evaluate the following integrals or state that they diverge. $$\int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x$$

Short Answer

Expert verified
Answer: The value of the integral is $$2(e - 1).$$

Step by step solution

01

Check for Convergence

The integral has limits from 0 to 1. Let's examine the behavior of the integrand as x approaches 0. Since the denominator is \(\sqrt{x}\), it goes to infinity when x goes to 0, and the exponential function in the numerator (\(e^{\sqrt{x}}\)) goes to 1. So the integrand behaves like \(\frac{1}{\sqrt{x}}\) near x=0. Since the integral of \(\frac{1}{\sqrt{x}}\) converges, we can conclude that the given integral converges as well.
02

Substitution Method

Now that we know that the integral converges, let's find its value. To do this, we will use the substitution method. Let $$u = \sqrt{x}.$$ Then, $$x = u^2$$ and $$d x = 2u du$$. Now, we'll perform the substitution: $$\int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = \int_{0}^{1} \frac{e^u}{u} (2u) du.$$
03

Simplify the Integral

Now, we will simplify the integral: $$\int_{0}^{1} \frac{e^u}{u} (2u) du = 2 \int_{0}^{1} e^u du.$$
04

Evaluate the Integral

It's easy to evaluate this integral directly: $$2 \int_{0}^{1} e^u du = 2 [e^u]_{0}^{1} = 2(e^1 - e^0) = 2(e - 1).$$ Thus, the value of the integral $$\int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x$$ is $$2(e - 1).$$

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Most popular questions from this chapter

The reaction of chemical compounds can often be modeled by differential equations. Let \(y(t)\) be the concentration of a substance in reaction for \(t \geq 0\) (typical units of \(y\) are moles/L). The change in the concentration of the substance, under appropriate conditions, is \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph and compare the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).

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