Problem 58 - Applied Calculus

Question

Use integration by substitution and the Fundamental Theorem to evaluate the definite integrals. $$\int_{-1}^{2} \sqrt{x+2} d x$$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{14}{3} \).

Step 1: Identify the Inner Function

Recognize the expression inside the square root as the inner function. Here, that function is \( u = x + 2 \).

Step 2: Determine the Differential

Calculate the differential of \( u \), giving \( du = dx \).

Step 3: Change the Limits of Integration

Substitute the limits of integration based on the new variable \( u \). When \( x = -1 \), \( u = -1 + 2 = 1 \). When \( x = 2 \), \( u = 2 + 2 = 4 \). The new limits are from 1 to 4.

Step 4: Rewrite the Integral

Replace \( x \) with \( u \) in the integral: \( \int_{1}^{4} \sqrt{u} \, du \).

Step 5: Integrate Using Power Rule

Use the power rule for integration: \( \int u^{1/2} \ du = \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2} \).

Step 6: Evaluate the Definite Integral

Apply the Fundamental Theorem of Calculus, evaluating \( \frac{2}{3}u^{3/2} \) from \( u=1 \) to \( u=4 \). First substitute 4: \( \frac{2}{3}(4)^{3/2} \), then substitute 1: \( \frac{2}{3}(1)^{3/2} \).

Step 7: Simplify and Subtract

Calculate \( \frac{2}{3}(4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3} \) and \( \frac{2}{3}(1)^{3/2} = \frac{2}{3} \times 1 = \frac{2}{3} \). The result is \( \frac{16}{3} - \frac{2}{3} = \frac{14}{3} \).

Key Concepts

Integration by Substitution

Integration by substitution is like unraveling a knot in calculus. It's a method to simplify integrals, especially when dealing with complex expressions. The main idea is to transform a complicated integral into a simpler one. To do this, you identify a portion of the expression as a new variable, usually labeled as \( u \). This transforms the integral into a potentially easier form.

In practice, you will:

• Identify the inner function which is suitable to set as \( u \). For example, in \( \sqrt{x+2} \), let \( u = x + 2 \).
• Find the differential: calculate \( du = dx \) from \( u = x + 2 \), which helps in replacing parts of the integral.
• Rewrite the original integral in terms of \( u \), changing from the variable \( x \) to \( u \).

Through these steps, the integral often becomes more straightforward, making it easier to solve.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges differential and integral calculus. It links the concept of differentiation with integration, giving a powerful way to evaluate definite integrals.

Here's how it works:

• After substituting and integrating, you express the antiderivative of the new function. For example, after integrating \( \int \sqrt{u} \, du \), find the antiderivative, which is \( \frac{2}{3} u^{3/2} \).
• Utilize the theorem to evaluate this antiderivative between your new limits of integration, turning an otherwise complex calculation into an arithmetic one. In our example, evaluate \( \frac{2}{3} u^{3/2} \) from \( u = 1 \) to \( u = 4 \).

The simplicity and beauty of this theorem come from its ability to make the calculation of definite integrals straightforward by focusing on antiderivatives.

Changing Limits of Integration

When using substitution, changing the limits of integration is essential. It ensures that the new integral corresponds to the same interval as the original. This adjustment aligns the integral's limits with the substitution used.

Steps involved include:

• Calculate new limits based on the substituted variable. For instance, when \( u = x + 2 \), substitute the original limits \( x = -1 \) and \( x = 2 \) to find \( u \)-values. For \( x = -1 \), \( u = 1 \). For \( x = 2 \), \( u = 4 \).
• Update the integral with these new limits, like changing from \( \int_{-1}^{2} \sqrt{x+2} \, dx \) to \( \int_{1}^{4} \sqrt{u} \, du \).

This change ensures the integrity of the integral while simplifying the computation in terms of the new variable.