Problem 24 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{1}^{4} \sqrt{u} d u$$

Step-by-Step Solution

Answer

The definite integral \(\int_{1}^{4} \sqrt{u} du\) can be evaluated by finding the antiderivative of \(\sqrt{u}\), which is \(\frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C\), and then applying the Fundamental Theorem of Calculus. The final answer is \(\dfrac{14}{3}\).

Step 1: Calculate Antiderivative of the Function

First, we need to find the antiderivative of \(\sqrt{u}\). We can rewrite this function as \(u^{\frac{1}{2}}\). Using the power rule for integration, we can find the antiderivative: \[ \int{u^{\frac{1}{2}}}du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C \]

Step 2: Fundamental Theorem of Calculus

Now, we will use the Fundamental Theorem of Calculus to evaluate the definite integral. The Fundamental Theorem of Calculus states that for a continuous function \(f(u)\) on an interval \([a,b]\), if \(F(u)\) is the antiderivative of \(f(u)\), then: \[\int_a^b f(u) du = F(b) - F(a)\] Using the antiderivative we found in Step 1, we can evaluate the definite integral: \[ \int_{1}^{4} \sqrt{u} d u = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \Big|_1^4 \] Now, we need to apply the limits of the integral: \[ \frac{4^{\frac{3}{2}}}{\frac{3}{2}} - \frac{1^{\frac{3}{2}}}{\frac{3}{2}} \] Simplifying this expression: \[ \frac{2}{3}(8 - 1) \]

Step 3: Final Answer

Now we have the final answer: \[ \int_{1}^{4} \sqrt{u} du = \frac{2}{3}(7) = \frac{14}{3} \] So the value of the definite integral is \(\dfrac{14}{3}\).

Key Concepts

Antiderivative

Calculating an antiderivative is a crucial step when tackling definite integrals. Think of an antiderivative as the opposite of taking a derivative. While derivatives measure change, antiderivatives help us find original functions or expressions.
When you see a function like \\( \sqrt{u} \), it can be challenging to differentiate but rewriting it as \\( u^{\frac{1}{2}} \) simplifies the process. This step enables us to apply familiar mathematical rules.
If you integrate \\( u^{\frac{1}{2}} \), you use the integration power rule: increase the exponent by one and divide by the new exponent. It transforms to:

• \( \int{u^{\frac{1}{2}}}du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C \)

This expression now represents the family of functions whose derivative is \\( u^{\frac{1}{2}} \). The constant \\( C \) reminds us that there are infinitely many antiderivatives differing by a constant.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation and integration. It introduces a simple method for calculating a definite integral once we know an antiderivative.
The theorem states that if \\( F(u) \) is an antiderivative of \\( f(u) \) on a continuous interval \\([a, b]\), evaluating the definite integral involves:

• \( \int_a^b f(u) du = F(b) - F(a) \)

This means applying the upper limit (\( b \)) and the lower limit (\( a \)) to the antiderivative you found. For the problem, this becomes:

• \( \int_{1}^{4} \sqrt{u} du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \Big|_1^4 \)

Plugging in the values of the limits brings us to an answer, showcasing how integration can help determine the area under a curve between two points.

Integration Power Rule

The integration power rule is a versatile tool for finding antiderivatives of power functions. Whenever you see a variable raised to a power, you can use this rule.
The rule states: Raise the exponent of the function by one and divide the term by this new exponent. For instance, to integrate \\( u^{n} \), this becomes:

• \( \int u^n du = \frac{u^{n+1}}{n+1} + C \)

Applying this rule to \\( u^{\frac{1}{2}} \) gives:

• Add 1 to the exponent \\( \left(\frac{1}{2} + 1 = \frac{3}{2}\right) \).
• Divide the term by \\( \frac{3}{2} \), resulting in \\( \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \).

With practice, the integration power rule becomes a straightforward mechanism for dealing with various power functions, making definite integrals and derivatives more manageable.