Problem 41 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: $$\int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = \int_{0}^{1} \frac{e^u}{u} (2u) du.$$

Step 3: Simplify the Integral

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

Step 4: 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 $$2(e - 1).$$