Problem 16 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{1}^{18} \sqrt{\frac{3}{z}} d z\)

Step-by-Step Solution

Answer

The integral evaluates to \(6\sqrt{6} - 2\sqrt{3}\).

Step 1: Simplify the integrand

First, we need to simplify the integrand. The expression \( \sqrt{\frac{3}{z}} \) can be rewritten using fractional exponents as \( 3^{\frac{1}{2}} \cdot z^{-\frac{1}{2}} \). Hence, the integral becomes: \[ \int_{1}^{18} 3^{\frac{1}{2}} z^{-\frac{1}{2}} \, dz \]

Step 2: Factor out the constant

Since \( 3^{\frac{1}{2}} \) is a constant, factor it out of the integral:\[ 3^{\frac{1}{2}} \int_{1}^{18} z^{-\frac{1}{2}} \, dz \]

Step 3: Integral of power function

To integrate \( z^{-\frac{1}{2}} \), recall the rule for integrating power functions: \[ \int z^n \, dz = \frac{z^{n+1}}{n+1} + C \]For \( n = -\frac{1}{2} \), the integral becomes:\[ \frac{z^{\frac{1}{2}}}{\frac{1}{2}} = 2z^{\frac{1}{2}} \]

Step 4: Apply limits of integration

Substitute the limits of integration into the evaluated antiderivative:\[ 3^{\frac{1}{2}} \left[ 2z^{\frac{1}{2}} \right]_{1}^{18} \]. Calculate:\[ 3^{\frac{1}{2}} \left( 2 \times 18^{\frac{1}{2}} - 2 \times 1^{\frac{1}{2}} \right) \]

Step 5: Perform calculations

Evaluate the expression:- Compute \( 18^{\frac{1}{2}} = \sqrt{18} = 3\sqrt{2} \), and- \( 1^{\frac{1}{2}} = 1 \).Thus, the expression becomes\[ 3^{\frac{1}{2}} \left( 2 \times 3\sqrt{2} - 2 \times 1 \right) \]Simplify to\[ 3^{\frac{1}{2}} \times (6\sqrt{2} - 2) \]

Step 6: Final computation

Calculate the final expression:- Since \( 3^{\frac{1}{2}} = \sqrt{3} \), the expression becomes:\[ \sqrt{3} \cdot (6\sqrt{2} - 2) \]Distribute to finish the computation:\[ 6\sqrt{6} - 2\sqrt{3} \]

Step 7: Conclusion

The evaluated integral \[ \int_{1}^{18} \sqrt{\frac{3}{z}} \, dz \] is \[ 6\sqrt{6} - 2\sqrt{3} \]

Key Concepts

Fractional Exponents

Fractional exponents offer an alternative to radical expressions, making calculations easier and neater. For example, the square root of a number, sometimes written \( \sqrt{x} \), can instead be expressed as \( x^{\frac{1}{2}} \). This notation relates directly to the concept of raising a number to a power, which is helpful in integration. When dealing with the expression \( \sqrt{\frac{3}{z}} \), it can be rewritten as \( 3^{\frac{1}{2}} \cdot z^{-\frac{1}{2}} \). Here:

• The fractional exponent \( \frac{1}{2} \) indicates taking the square root.
• The negative exponent \( -\frac{1}{2} \) signifies the reciprocal, as in \( \frac{1}{z^{\frac{1}{2}}} \).

By rewriting square roots and radicals using fractional exponents, you can efficiently simplify integrals or other algebraic expressions.

Power Rule for Integration

The power rule for integration is one of the simplest and most used formulas when solving integrals. It helps in finding antiderivatives for functions like \( x^n \). The rule states:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]This tells us that to integrate a power function, you should:

• Add 1 to the exponent.
• Divide by the new exponent.

In our exercise, we have the expression \( z^{-\frac{1}{2}} \), which fits perfectly with the rule. Using the power rule, we integrate to find:\[\frac{z^{\frac{1}{2}}}{\frac{1}{2}} = 2z^{\frac{1}{2}} \]This calculation highlights how simply applying this rule allows us to solve integrals of functions with fractional exponents.

Limits of Integration

Limits of integration define the interval over which we compute the definite integral. In this problem, the function \( \sqrt{\frac{3}{z}} \) is integrated over the interval \[ [1, 18] \].Here's what to do:

• Find the antiderivative of the function within these limits.
• Substitute the upper limit into the antiderivative and compute.
• Substitute the lower limit and compute.
• Subtract the value of the lower limit from the value of the upper limit.

Therefore, if \( F(z) \) is the antiderivative, the definite integral is\[F(18) - F(1)\]This process helps find the area under the curve between the two points.

Simplifying Integrands

Simplifying integrands involves rearranging terms or factoring constants out of an integral to make it more straightforward and manageable to compute. Consider the integrand \( \sqrt{\frac{3}{z}} \). At first glance, it may seem a bit tricky. By rewriting it in terms of fractional exponents, we get \( 3^{\frac{1}{2}} \cdot z^{-\frac{1}{2}} \), which is already simplified to a better form for integration. Here's why it helps:

• It turns a radical expression into an algebraic one.
• Allows factoring out constants like \( 3^{\frac{1}{2}} \) to simplify calculations further.

Factoring out the constant simplifies our work to focus on integrating \( z^{-\frac{1}{2}} \) alone, which is significantly easier!