Problem 14 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

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

Step-by-Step Solution

Answer

The definite integral \(\int_{1}^{4} x \sqrt{x+1} d x\) is equal to \(\frac{170}{3}\) after applying substitution, evaluating antiderivatives, and using the Fundamental Theorem of Calculus.

Step 1: Find an antiderivative of the function

We need to first find an antiderivative of the function \(x\sqrt{x+1}\). We will perform a substitution, where we let \(u = x+1\), and \(d u = d x\). This means \(x = u - 1\). The integral will then transform into: \[ \int(u-1)\sqrt{u} du \]

Step 2: Evaluate the transformed integral

Next, we need to evaluate the transformed integral: \[ \int(u-1)\sqrt{u} du = \int u\sqrt{u} du - \int \sqrt{u} du \] Let's evaluate both these integrals: For \(\int u\sqrt{u} du\), we can rewrite it as: \[ \int u^{\frac{3}{2}} du = \frac{2}{5} u^{\frac{5}{2}} + C_1 \] For \(\int \sqrt{u} du\), we can rewrite it as: \[ \int u^{\frac{1}{2}} du = \frac{2}{3} u^{\frac{3}{2}} + C_2 \] Now we can combine both antiderivatives: \[ \frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}} + C \] Where \(C = C_1 - C_2\).

Step 3: Apply the Fundamental Theorem of Calculus

Now, we can substitute back for x and evaluate the definite integral over the interval [1, 4]. We have: \[ \int_{1}^{4} x\sqrt{x+1} dx =\left[\frac{2}{5} (x+1)^{\frac{5}{2}} - \frac{2}{3} (x+1)^{\frac{3}{2}}\right]_{1}^{4} \] Evaluating at the limits of integration: \[ \bigg(\frac{2}{5}(4+1)^{\frac{5}{2}} - \frac{2}{3}(4+1)^{\frac{3}{2}}\bigg) - \bigg(\frac{2}{5}(1+1)^{\frac{5}{2}} - \frac{2}{3}(1+1)^{\frac{3}{2}}\bigg) \]

Step 4: Compute the final answer

Finally, we compute the answer: \[ \bigg(\frac{2}{5}(5)^{\frac{5}{2}} - \frac{2}{3}(5)^{\frac{3}{2}}\bigg) - \bigg(\frac{2}{5}(2)^{\frac{5}{2}} - \frac{2}{3}(2)^{\frac{3}{2}}\bigg) = \frac{170}{3} \] So, the value of the definite integral is: \[ \int_{1}^{4} x \sqrt{x+1} dx = \frac{170}{3} \]

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve integrals that are difficult to evaluate otherwise.

• This method involves changing the variable of integration to simplify the given function.
• The new variable should ideally turn the integral into something more manageable.

In the given exercise, we performed a substitution where we set \( u = x + 1 \) and \( du = dx \). This effectively transforms the integral from \( \int_{1}^{4} x \sqrt{x+1} \, dx \) into \( \int (u-1) \sqrt{u} \, du \).
By making these substitutions, we're expressing \( x \) in terms of \( u \) (i.e., \( x = u - 1 \)), simplifying the power of the expression inside the integral. This allows us to apply more straightforward integration techniques on the new function with respect to \( u \).Understanding and setting up the right substitution requires practice, but it is invaluable as it can drastically simplify the integration process.
The primary goal is to reduce the complexity of the expression to a familiar form, such as a polynomial of a simple power.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus plays a crucial role in connecting derivatives to integrals.

• This theorem is what allows us to evaluate the definite integral once we have an antiderivative.
• It states that if a function is continuous over a certain interval, then the definite integral over that interval is simply the difference between the values of an antiderivative at the endpoints of the interval.

In our problem, after determining the antiderivative \( \frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}} + C \), we used the theorem to evaluate it over the interval [1, 4].
The theorem allowed us to substitute back \( x+1 \) for \( u \) and evaluate:\[\int_{1}^{4} x \sqrt{x+1} \, dx = \left[ \frac{2}{5} (x+1)^{\frac{5}{2}} - \frac{2}{3} (x+1)^{\frac{3}{2}} \right]_{1}^{4}\]Thus, providing a mechanism to compute the definite integral by plugging in the boundary values. An understanding of this theorem is pivotal as it bridges the conceptual gap between differential calculus (derivatives) and integral calculus.

Antiderivative

Finding the antiderivative, also known as the indefinite integral, is a fundamental part of integral calculus.

• An antiderivative of a function is another function whose derivative is the original function.
• It's the reverse operation of taking a derivative.

For our problem, once we substituted, we needed to find antiderivatives of \( u^{\frac{3}{2}} \) and \( u^{\frac{1}{2}} \).

• For \( \int u^{\frac{3}{2}} \, du \), the antiderivative is \( \frac{2}{5} u^{\frac{5}{2}} + C_1 \).
• For \( \int u^{\frac{1}{2}} \, du \), it is \( \frac{2}{3} u^{\frac{3}{2}} + C_2 \).

We then combined these results to form a complete antiderivative that we used to apply the Fundamental Theorem of Calculus.
Understanding how to find antiderivatives is key to solving many integral problems, especially those involving definite integrals where we want to find the area under a curve.