Problem 57 - Calculus for Biology and Medicine

Question

Use substitution to evaluate the definite integrals. $$ \int_{1}^{9} \frac{1}{\sqrt{x}} e^{-\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The definite integral equals \( 2e^{-1} - 2e^{-3} \).

Step 1: Choose a Substitution

To simplify the integral, choose a substitution. Let \( u = \sqrt{x} \). Then, \( x = u^2 \) and \( dx = 2u \, du \).

Step 2: Change the Limits of Integration

Substitute the current limits for \( x \) to find the new limits for \( u \): when \( x = 1 \), \( u = \sqrt{1} = 1 \); when \( x = 9 \), \( u = \sqrt{9} = 3 \). So, the limits for \( u \) are from 1 to 3.

Step 3: Rewrite the Integral in Terms of u

Substitute \( u \) and \( dx \) into the integral: \[ \int_{1}^{9} \frac{1}{\sqrt{x}} e^{-\sqrt{x}} \, dx = \int_{1}^{3} \frac{1}{u} e^{-u} (2u \, du) = \int_{1}^{3} 2 e^{-u} \, du \].

Step 4: Integrate with Respect to u

Integrate \( \int 2 e^{-u} \, du \). The integral of \( e^{-u} \) is -\( e^{-u} \), so the integral becomes: \[ 2 \int e^{-u} \, du = -2 e^{-u} \].

Step 5: Evaluate the Definite Integral

Evaluate the definite integral from 1 to 3: \[-2e^{-u} \bigg|_{1}^{3} = [-2e^{-3}] - [-2e^{-1}] = 2e^{-1} - 2e^{-3} \].

Step 6: Simplify the Result

Simplify the expression: \( 2e^{-1} - 2e^{-3} = 2\left( \frac{1}{e} - \frac{1}{e^3} \right) \). The result is in its simplest form.

Key Concepts

Substitution Method

The Substitution Method is a powerful tool for evaluating integrals, especially when dealing with more complex functions. In this technique, we change the variable of integration to simplify the integral into a more manageable form.
By choosing a substitution:

• Letting \( u = \sqrt{x} \) transforms \( x = u^2 \), which simplifies the function inside the integral.
• Finding the derivative, \( dx = 2u \, du \), allows the expression to be rewritten in terms of \( u \).

This substitution method turns a difficult integration problem into a simpler one by altering both the function and its derivatives. It essentially re-maps the problem so we can apply more straightforward integration techniques.By setting new limits and changing the differential, the resulting integral in terms of \( u \) is often easier to solve, as demonstrated in our specific example here.

Integration Techniques

Integration Techniques are strategies we use to solve integrals effectively. The substitution method is just one, but there are many others, each suited to different kinds of functions.
With substitution complete, the new integral was:

• \( \int_{1}^{3} 2 e^{-u} \, du \)

This can be integrated using the basic rules of exponential functions. The integral of \( e^{-u} \) is straightforward: it's \(-e^{-u} \). By applying this knowledge:

• The integration becomes \( -2 e^{-u} \), reflecting the constant of multiplication on the outside before integration.

Mastering such methods increases problem-solving efficiency. Different integrals require different techniques, but understanding when and how to apply these is key to success in calculus. Substitutions and simplifications help transform integrals into forms that are easier to solve, like converting them into basic exponential integrals.

Limits of Integration

Limits of Integration are another important aspect when dealing with definite integrals. These limits define where the integration starts and ends on the x-axis. Changing variables using substitution also requires changing these limits to match the new variable.
In our solved example:

• Original limits were \( x = 1 \) to \( x = 9 \).
• Substituting \( u = \sqrt{x} \) changed the limits to \( u = 1 \) to \( u = 3 \).

This ensures the integral is correctly evaluated over the new domain of the function, preserving the area under the curve that the original problem defines.
Correctly adjusting limits is crucial. It ensures that the transformation aligns the integral's evaluation with its original scope, allowing us to compute a meaningful and accurate result. In our case, the adjusted integral gives us the difference \( 2e^{-1} - 2e^{-3} \), representing the originally defined area in a newly transformed equation.