Problem 11 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Evaluate the double integrals.\(\int_{0}^{4} \int_{0}^{\sqrt{x}} x^{2} y d y d x\)

Step-by-Step Solution

Answer

The value of the double integral is 32.

Step 1: Set up the double integral

The given double integral is \[\begin{equation}\begin{aligned}\begin{split}\int_{0}^{4} \int_{0}^{\sqrt{x}} x^{2} y d y d x.\end{split}\end{aligned}\end{equation}\]Identify the order of integration as from inner to outer (\text{inner: } dy, \text{outer: } dx).

Step 2: Integrate with respect to y

Integrate the inner integral \[\begin{equation}\int_{0}^{\sqrt{x}} x^{2} y d y.\end{equation}\]Since x is treated as a constant, we can apply the power rule for integration: \[\begin{equation}x^{2} \int_{0}^{\sqrt{x}} y d y.\end{equation}\]The integral of y is \[\begin{equation}\frac{y^2}{2}\end{equation}\] evaluated from 0 to \[\begin{equation}\sqrt{x}.\end{equation}\]Thus, we'll evaluate the integral as follows: \[\begin{equation}x^{2}\left[\frac{y^2}{2}\right]_{0}^{\sqrt{x}} = x^2\left[\frac{x}{2}\right] = x^2 \cdot \frac{x}{2} = \frac{x^3}{2}.\end{equation}\]

Step 3: Integrate with respect to x

Now, integrate the result with respect to x: \[\begin{equation} \int_{0}^{4} \frac{x^3}{2} d x \end{equation}\]. Factor out the constant \( \frac{1}{2} \): \[\begin{equation} \frac{1}{2} \int_{0}^{4} x^3 d x. \end{equation}\]. Apply the power rule for integration: \[\begin{equation} \int_{0}^{4} x^3 d x = \frac{x^4}{4} \end{equation}\] evaluated from 0 to 4. So, \[\begin{equation} \frac{1}{2} \cdot \left[ \frac{x^4}{4} \right]_{0}^{4} = \frac{1}{2} \cdot \left[ \frac{4^4}{4} - \frac{0^4}{4} \right] = \frac{1}{2} \cdot \left[ \frac{256}{4} \right] = \frac{1}{2} \cdot 64 = 32. \end{equation}\]

Key Concepts

integration order

When working with double integrals, the order of integration is a crucial first step. In this exercise, we start by integrating with respect to the inner variable, denoted as \( y \), and then proceed to the outer variable, \( x \). This order is indicated in the integral expression: \(\int_{0}^{4} \int_{0}^{\sqrt{x}} x^{2} y \ d y \ d x\). Here, integrating 'dy' first means we consider \( x \) as a constant while solving the inner integral. This structured process ensures that we handle variables appropriately at each integration step.

power rule for integration

The power rule for integration states that \( \int x^n \ dx = \frac{x^{n+1}}{n+1}\ \), where \( n \) is any real number other than -1. This rule simplifies integration of polynomials. In our exercise, we first apply it to the inner integral, treating \( x \) as a constant. The inner integral, \( x^{2} \int_{0}^{\sqrt{x}} y \ dy \), uses the power rule: \( \fra\{y^2}{2} \). Evaluating from 0 to \( \sqrt{x} \), we simplify this expression to \( x^{2} \ * \ \fra\{x}{2} = \fra\{x^3}{2} \). We then apply the power rule again to the outer integral: \( \fra\{1}{2} \int_{0}^{4} x^3 \ dx = \fra\{1}{2} \ * \ \fra\{x^4}{4} \), evaluated from 0 to 4, simplifying to a final result.

evaluating definite integrals

Evaluating definite integrals involves computing values within specific bounds. After integrating, we replace the variable with these bounds, subtracting the lower bound evaluation from the upper one. For the given exercise, integrating the inner integral within limits 0 to \( \sqrt{x} \) gives \( x^{2} \frac{y^2}{2} \bigg|_{0}^{\sqrt{x}} = x^2 \frac{x}{2} = \fra\{x^3}{2} \). Next, the outer integral evaluates within 0 to 4 as: \(\fra\{1}{2} \int_{0}^{4} x^3 \ dx\), resulting in \(\fra\{1}{2} \frac{x^4}{4} \bigg|_{0}^{4} = \fra\{1}{2} \frac{256}{4} = 32\). This step-by-step process streamlines to our solution of 32 after accounting for given bounds.