Problem 44 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

The definite integral evaluates to \(2\sqrt{2} - 2 \sqrt{e^{-1}+1}\)

Step 1: Identify Substitution

Look for parts of the integral that might simplify with a substitution. In this case, if you let \(u = e^{-x}+1\), its derivative can eliminate the \(e^{-x}\) in the numerator after substitution.

Step 2: Change of Variables

Using the substitution \(u = e^{-x}+1\), we find \(-du = e^{-x}\, dx\). Substituting these into the integral and changing the limits (when \(x=0\), \(u=2\); when \(x=1\), \(u=e^{-1}+1\)), the integral becomes: \(-\int_{2}^{e^{-1}+1} \frac{1}{\sqrt{u}} du\)

Step 3: Integrate Using Power Rule

The integral of \(\frac{1}{\sqrt{u}}\) can be rewritten as \(u^{-1/2}\) and integrated using the power rule to get \(-2\sqrt{u}\). Evaluated between the limits \(e^{-1}+1\) and 2, the integral becomes: \(-2[\sqrt{e^{-1}+1}-\sqrt{2}]\)

Step 4: Simplify

After simplification, the result is: \(2\sqrt{2} - 2 \sqrt{e^{-1}+1}\)

Key Concepts

U-Substitution

Understanding u-substitution is key to solving a range of integration problems, especially when encountering a composite function that doesn't fit standard integration formulas. The goal of u-substitution is to simplify an integral, making it easier to solve by changing the integral's variable and differential component (usually written as 'dx').

Imagine you're given a complex function to integrate, much like trying to fit a square peg into a round hole; it just won't work with your standard integration techniques. This is where u-substitution comes in—it's like shaping the peg so it perfectly fits the hole. In the exercise, the choice of u being e^{-x} + 1 strategically molds the function into a simpler form that fits within the confines of basic integration rules by removing the extraneous e^{-x} component.

It is helpful to think of u-substitution as a two-step dance: first, identifying the correct substitution (the dance move), and second, adjusting the integral's limits to match this new variable (changing the dance tempo). This way, you ensure that the function does not just look easier but is genuinely aligned with the standard integrals you know how to solve.

Integration Power Rule

The integration power rule is one of the most fundamental rules in calculus. It's your bread and butter when handling integrals involving powers of x. When faced with an expression of the form x^n, the integration power rule is the reliable tool to reach for.

According to this rule, to integrate an expression like x^n, you increment the exponent by 1 to get x^{n+1} and then divide by this new exponent, yielding \( \frac{1}{n+1}x^{n+1} \). Remember, this rule only applies when n is not equal to -1 since that leads to a logarithmic function rather than a power function.

In our exercise, after using u-substitution, we encountered an integral in the form of \( u^{-1/2} \), which falls right into the lap of the integration power rule. Applying this rule transformed the integral into a more manageable -2\sqrt{u}, moving us one step closer to finding the definite integral's value. Just think of the power rule as the simplifying factor that turns a wild algebraic expression into a tame one that's ready to offer up its integral without a fight.

Change of Variables

The technique of change of variables in definite integrals is a powerful ally for calculus students. When the regular x's and y's start to look daunting, a change of variables can be the lifeline that rescues you from the complexity of an integral.

In essence, changing variables is like changing perspective. It's about transforming the integral into a new form that showcases its simpler, more approachable side. It's not just about swapping letters; it's about creating a bridge between the given problem and a solution you know how to find. In our definite integral exercise, the change of variables involved replacing e^{-x} and dx with -du and accordingly adjusting the integral's limits to reflect the new variable u, which transitioned seamlessly from a challenging integral into a form that practically solves itself.

Always remember to adjust the limits of integration when you perform a change of variables in a definite integral; this ensures the solution stays accurate within the new variable's domain. It's like being careful to use the right map scale when you shift your geographical focus—otherwise, you risk getting lost on the path to the right answer.