Problem 24 - Calculus Single Variable

Question

Use the method of substitution to evaluate the definite integrals. $$ \int_{0}^{5} \sqrt{1+3 x} d x $$

Step-by-Step Solution

Answer

The evaluated integral equals 42.

Step 1: Determine the Substitution

To use substitution, set a new variable to simplify the expression inside the integral. Let \( u = 1 + 3x \). This means the expression inside the square root becomes \( \sqrt{u} \).

Step 2: Find the Derivative

Differentiate the substitution equation with respect to \( x \) to find \( du/dx \). Doing so gives \( \frac{du}{dx} = 3 \), or equivalently \( dx = \frac{du}{3} \).

Step 3: Substitute Variables in the Integral

Substitute \( u \) and \( dx \) back into the integral. The new integral becomes \( \int \sqrt{u} \cdot \frac{du}{3} \). Simplify to \( \frac{1}{3} \int \sqrt{u} \, du \).

Step 4: Change the Limits of Integration

The original limits are \( x = 0 \) and \( x = 5 \). Substitute these into \( u = 1 + 3x \) to find the new limits: when \( x = 0 \), \( u = 1 \); when \( x = 5 \), \( u = 16 \). So the limits change from 0-5 to 1-16.

Step 5: Evaluate the Integral

Compute \( \frac{1}{3} \int_{1}^{16} u^{1/2} \, du \). Integrate \( u^{1/2} \) with respect to \( u \) to get \( \frac{2}{3} u^{3/2} \). Evaluate this from \( u = 1 \) to \( u = 16 \) to find: \( \frac{2}{3} \left[u^{3/2}\right]_{1}^{16} \).

Step 6: Calculate the Result

Substitute the limits into the evaluated expression: \( \frac{2}{3} \times 16^{3/2} - \frac{2}{3} \times 1^{3/2} \). Compute \( 16^{3/2} = 64 \) and \( 1^{3/2} = 1 \). So the result is \( \frac{2}{3} (64 - 1) = \frac{2}{3} \times 63 = 42 \).

Key Concepts

Substitution Method

The substitution method, sometimes known as "u-substitution," is a technique used to simplify the process of integration. This method is particularly useful for definite integrals where the integrand can be transformed into a simpler function. The key idea is to set a part of the integrand equal to a new variable, often denoted by \( u \). In our problem, we set \( u = 1 + 3x \). This new variable \( u \) helps to streamline the integration process by simplifying complex expressions. Once \( u \) is chosen, we need to find its derivative with respect to \( x \), represented as \( \frac{du}{dx} \). This step is crucial because it allows us to express \( dx \) in terms of \( du \), completing the transformation. Here, \( \frac{du}{dx} = 3 \), leading to \( dx = \frac{du}{3} \).Using substitution, the original integral becomes more manageable: \( \sqrt{1+3x} \to \sqrt{u} \). By substituting \( u \) and \( dx \), the integral changes to \( \frac{1}{3} \int \sqrt{u} \, du \). This transformation is what makes the integral easier to evaluate. By turning a complex expression into something simpler, integration becomes more straightforward.

Limits of Integration

In definite integrals, limits of integration indicate the bounds between which we evaluate the function. When applying the substitution method, these limits must also be transformed, or updated, to correspond with the new variable \( u \). This is done to ensure that the evaluation correctly reflects the area under the curve for the transformed integral.Our initial integral is bounded between \( x=0 \) and \( x=5 \). When transforming these limits using the equation \( u = 1 + 3x \), we substitute the limits into this expression:

• For \( x = 0 \): \( u = 1 + 3(0) = 1 \)
• For \( x = 5 \): \( u = 1 + 3(5) = 16 \)

Thus, the new limits of integration for our transformed integral are \( u = 1 \) to \( u = 16 \). Updating the limits is essential to accurately compute the definite integral after making a substitution. It ensures continuity and accuracy in the area calculation of the transformed function.

Integral Evaluation

Transforming the integral helps us to evaluate it more easily. After substituting and adjusting limits, we are ready to evaluate the integral. Our transformed integral \( \frac{1}{3} \int_{1}^{16} \sqrt{u} \, du \) is now much simpler to work with.We focus on integrating \( \sqrt{u} \) with respect to \( u \). Converting \( \sqrt{u} \) to a power of \( u \), we have \( u^{1/2} \). The integral of \( u^{1/2} \) is \( \frac{2}{3}u^{3/2} \). This approach allows the use of basic power rule for integration, making it straightforward.The actual evaluation involves substituting the limits of integration back into the integral. With limits \( 1 \) and \( 16 \), we compute:\[ \frac{2}{3} \left[ u^{3/2} \right]_{1}^{16} = \frac{2}{3} (16^{3/2} - 1^{3/2}) \]Calculations show:

• \( 16^{3/2} = 64 \)
• \( 1^{3/2} = 1 \)

Substituting these values results in \( \frac{2}{3} (64 - 1) = \frac{2}{3} \times 63 = 42 \). The ease of evaluation exemplifies the power of transforming integrals using substitution.

Change of Variables

The change of variables is a mathematical technique that redefines an integral's variable and adjusts the expression accordingly. This method transforms the integral into a form that is often easier to solve. In this task, the original variable \( x \) is redefined as \( u = 1 + 3x \). This not only simplified the integrand but also allowed for a smoother integration process by reducing a complex expression into \( \sqrt{u} \).Changing variables involves finding the derivative of the new variable with respect to the old one, which guides the setup for replacement. Here, \( \frac{du}{dx} = 3 \) allowed us to find \( dx = \frac{du}{3} \). Such transformations result in changes in both the integrand and the differentials.By substituting, the integral's complexity is reduced, turning it into something far more manageable: \( \frac{1}{3} \int \sqrt{u} \, du \). This showcases how changing variables can transform challenging integrals, simplifying both the integration and subsequent evaluation steps.