Question

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{0}^{4} \int_{0}^{-x / 2+2}\left(3 x^{2}-y+2\right) d y d x $$

Short answer

The evaluated integral is \( \frac{208}{3} \). The reversed order integral is \( \int_{0}^{2} \int_{2y}^{4} (3x^2 - y + 2) \, dx \, dy \).

Step-by-step solution

Evaluate the Inner Integral

First, we need to evaluate the inner integral \( \int_{0}^{-x / 2 + 2} (3x^2 - y + 2) \, dy \). We take the antiderivative of \( (3x^2 - y + 2) \) with respect to \( y \), which is \( 3x^2y - \frac{y^2}{2} + 2y \).

Evaluate the Inner Integral with Bounds

Next, evaluate the antiderivative from Step 1 at the bounds \( 0 \) and \( -x/2 + 2 \). This involves substitution: \[ \left[ 3x^2(-\frac{x}{2} + 2) - \frac{(-\frac{x}{2} + 2)^2}{2} + 2(-\frac{x}{2} + 2) \right] - \left[ 0 \right] \]. Simplify the expression to get \( 5x^2 - \frac{5x}{2} + 4 \).

Evaluate the Outer Integral

Now, evaluate the outer integral \( \int_{0}^{4} (5x^2 - \frac{5x}{2} + 4) \, dx \). Find the antiderivative with respect to \( x \): \[ \frac{5x^3}{3} - \frac{5x^2}{4} + 4x \]. Evaluate this antiderivative from \( 0 \) to \( 4 \).

Apply the Bounds to the Antiderivative

Substitute \( 4 \) and \( 0 \) into the expression from Step 3: \[ \left( \frac{5(4)^3}{3} - \frac{5(4)^2}{4} + 4(4) \right) - \left( 0 \right) \]. Compute to get the result \( \frac{208}{3} \).

Change the Order of Integration

To rewrite the integral with the order of integration reversed, examine the bounds. The original integral\( \int_{0}^{4} \int_{0}^{-x/2+2} \, dy \, dx \) suggests: 1. \( 0 \leq x \leq 4 \)2. \( 0 \leq y \leq -x/2 + 2 \).For reversed integration, consider fixing \( y \) first: \( 0 \leq y \leq 2 \) and \( 2y \leq x \leq 4 \). Thus, the integral becomes \( \int_{0}^{2} \int_{2y}^{4} \left( 3 x^{2} - y + 2 \right) \ dx \, dy \).

Key concepts

Order of Integration

The order of integration in iterated integrals plays a crucial role in how we tackle solving these integrals. Simply put, it dictates which variable you integrate with respect to first. This decision impacts the bounds of the integration you choose for each variable.

In our original exercise, the order of integration has us integrating with respect to \( y \) first and then \( x \). This means our inner integral is evaluated concerning \( y \), using the bounds dependent on \( x \): from \( 0 \) to \(-x/2 + 2\). After evaluating the inner integral, we take its result and integrate it with respect to \( x \) over the bounds from \( 0 \) to \( 4 \).

Changing the order of integration means potentially simplifying the computation process or tackling constraints more effectively. When reversing the order, as done in the exercise solution, we first fix the upper and lower limits of \( y \) (from \( 0 \) to \( 2 \)). The inner integral, then in \( x \), changes its bounds to accommodate \( y \) and runs from \( 2y \) to \( 4 \). This swap sometimes makes integration easier or more suited to particular types of functions.

When faced with a particularly cumbersome integral, adjusting the order of integration can be a strategic choice that simplifies the integration process, or even a necessity to express or solve the integral in a workable form.

Antiderivatives

Antiderivatives are fundamental in integration, serving as the reverse operation of differentiation. Simply put, finding an antiderivative of a function means identifying a function whose derivative gives the original function. This "reverse differentiation" forms the basis for calculating both definite and indefinite integrals.

In our task, we first find the antiderivative of the function with respect to \( y \). Given the integrand \( 3x^2 - y + 2 \), integrating it with respect to \( y \) results in the antiderivative: \( 3x^2y - \frac{y^2}{2} + 2y \). Each segment of the function is integrated individually:

• \( 3x^2 \) is treated as a constant in terms of \( y \), giving \( 3x^2y \).
• Integrating \( -y \) results in \( -\frac{y^2}{2} \).
• Integrating \(2\) with respect to \( y \) gives \(2y\).

The meticulous application of these rules enables us to evaluate integrals by calculating the antiderivatives, a process crucially reflective in both the initial exercise and projects into evaluating definite integrals overall.

Definite Integrals

Definite integrals allow us to compute the 'net area' under a curve, between two specific points. They are integral in calculating physical quantities like area, volume, displacement, etc., where a specific range is defined.

In iterated integrals, as seen in the exercise, after obtaining the antiderivative of a function, the next significant step is to evaluate it between specific bounds. This process includes substituting these bounds into the antiderivative and calculating the values to find the definite integral.

In our scenario, after finding the antiderivative of the inner integral, the bounds \( 0 \) and \( -x/2 + 2 \) were applied, leading to the expression \(5x^2 - \frac{5x}{2} + 4\). This expression itself then becomes an integrand for the outer integral.

Finally, the result of the outer integral evaluation, after substituting the bounds \(0\) and \(4\) back into the respective antiderivative \(\frac{5x^3}{3} - \frac{5x^2}{4} + 4x\), secures the definite integral, showing a calculated "area" or quantity represented by the mathematical operation.

Upper and Lower Bounds

Understanding upper and lower bounds is key to properly evaluating a definite integral. Bounds determine the scope of integration, dictating the values over which integration occurs.

In the given problem, bounds are specified for both variables. Initially, for the inner integral in terms of \( y \), the bounds depend on \( x \)—going from \( 0 \) to \(-x/2 + 2 \). For the outer integral, \( x \) goes from \( 0 \) to \( 4 \).

Reversing the order of integration also involves re-evaluating these bounds. When revised, the bounds for \( y \) are given as fixed values between \( 0 \) and \( 2 \), while the bounds for \( x \) now depend on \( y \), ranging from \( 2y \) to \( 4 \). Each change respects the geometric region defined by the original problem.

Recognizing and correctly manipulating these bounds ensures that our integration fully captures the region or quantity of interest designated by the integral—precision in setting these defines the correctness of the final result.