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Chapter 37: Problem 3

Evaluate \(\int_{-2}^{1} \int_{x^{2}+4 x}^{3 x+2} \mathrm{~d} y \mathrm{~d} x\)

Short Answer

Expert verified
The value of the double integral \(\int_{-2}^{1} \int_{x^{2}+4 x}^{3 x+2} \mathrm{~d} y \mathrm{~d} x\) is \(-\frac{1}{6}\).

Step by step solution

01

Integrate with respect to y

First, let's integrate the inner integral with respect to y: \[ \int_{x^2 + 4x}^{3x + 2}dy \] Since it is a simple integral with respect to y, we just add a y variable to the expression: \[ y\big|_{x^2 + 4x}^{3x + 2} \] Now, let's substitute the limits of integration: \[ (3x + 2) - (x^2 + 4x) = 2 - x^2 - x \]
02

Integrate the resulting expression with respect to x

Now, let's integrate the resulting expression with respect to x: \[ \int_{-2}^{1} (2 - x^2 - x) dx \] Using the power rule of integration, we have: \[ \int (2 - x^2 - x) dx = 2x - \frac{x^3}{3} - \frac{x^2}{2} + C \] Now, we will apply the limits of integration -2 and 1 to our result:
03

Apply the limits of integration -2 and 1

Substitute the upper limit (x=1) into the expression: \[ 2(1) - \frac{1^3}{3} - \frac{1^2}{2} + C = 2 - \frac{1}{3} - \frac{1}{2} \] Next, substitute the lower limit (x=-2) into the expression: \[ 2(-2) - \frac{(-2)^3}{3} - \frac{(-2)^2}{2} + C = -4 + \frac{8}{3} - 2 \] Now, subtract the result from the lower limit from the result of the upper limit to get the final answer: \[ 2 - \frac{1}{3} - \frac{1}{2} - (-4 + \frac{8}{3} - 2) = -\frac{1}{6} \] The value of the double integral is \(-\frac{1}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integration
Definite integration helps us find the total accumulated value of a function over a specific interval. Unlike an indefinite integral which represents a function family, a definite integral represents a number. In this exercise, we are evaluating a double integral, which accumulates values over a region in the xy-plane.

Here,
  • The function is integrated first with respect to \( y \), over a range defined by two functions of \( x \), \( x^2 + 4x \) and \( 3x + 2 \).
  • Next, it is integrated with respect to \( x \), from the lower bound \( -2 \) to the upper bound \( 1 \).
This process provides the entire accumulated output across the specified regions for both variables.
Limits of Integration
Limits of integration define the boundaries for the variable being integrated. They decide the region over which the function is evaluated. In a double integral, each variable will have its own set of limits.

For this problem, the limits for our inner integral in terms of \( y \) are variable instead of just constants:
  • Lower limit: \( y = x^2 + 4x \)
  • Upper limit: \( y = 3x + 2 \)
After integrating with respect to \( y \), we use numerical limits for \( x \):
  • Lower limit: \( x = -2 \)
  • Upper limit: \( x = 1 \)
These limits are then substituted back into the anti-derivative to calculate the definite integral value.
Power Rule of Integration
The power rule is a fundamental technique in integration that allows us to find the antiderivative of power functions easily. It is typically expressed as:
  • \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), given \( n eq -1 \).
In this exercise, the expression \( 2 - x^2 - x \) is integrated relative to \( x \).

Here's how we apply the power rule to each term:
  • The integral of \( 2 \) becomes \( 2x \) as it is a constant.
  • For \( -x^2 \), it becomes \( -\frac{x^3}{3} \).
  • For \( -x \), it becomes \( -\frac{x^2}{2} \).
By applying these rules, we achieve the antiderivative necessary to evaluate the definite integral.

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Most popular questions from this chapter

Evaluate \(\int_{0}^{\pi} \int_{0}^{\cos \theta} r \sin \theta \mathrm{d} r \mathrm{~d} \theta\)

A plane figure is bounded by the polar curve \(r=a(1+\cos \theta)\) between \(\theta=0\) and \(\theta=\pi\), and the initial line OA. Express as a double integral the first moment of area of the figure about OA, and evaluate the integral. If the area of the figure is known to be \(\frac{3 \pi a^{2}}{4}\) unit \(^{2}\), find the distance \((h)\) of the centroid of the figure from OA.

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