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Chapter 6: Problem 13

Evaluate. $$ \int_{0}^{2} \int_{0}^{x}\left(x+y^{2}\right) d y d x $$

Short Answer

Expert verified
The value of the double integral is 4.

Step by step solution

01

Understand the Integral

The given double integral is \( \int_{0}^{2} \int_{0}^{x} (x+y^2) \, dy \, dx \). This means we first integrate with respect to \( y \) while treating \( x \) as a constant, and then integrate the result with respect to \( x \).
02

Inner Integral with Respect to y

Integrate \( x+y^2 \) with respect to \( y \) from \( 0 \) to \( x \).\[\int_{0}^{x} (x + y^2) \, dy = \int_{0}^{x} x \, dy + \int_{0}^{x} y^2 \, dy \] Evaluate each separately:\(\int_{0}^{x} x \, dy = [xy]_0^x = x(x) - x(0) = x^2 \)\(\int_{0}^{x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^x = \frac{x^3}{3} - \frac{0^3}{3} = \frac{x^3}{3} \)Combine the results:\( x^2 + \frac{x^3}{3} \) is the result of the inner integration.
03

Outer Integral with Respect to x

Integrate the result from the previous step with respect to \( x \) from \( 0 \) to \( 2 \):\[\int_{0}^{2} \left( x^2 + \frac{x^3}{3} \right) \, dx = \int_{0}^{2} x^2 \, dx + \int_{0}^{2} \frac{x^3}{3} \, dx\]Evaluate each separately:\(\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \)\(\int_{0}^{2} \frac{x^3}{3} \, dx = \frac{1}{3}\left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{3} \left( \frac{16}{4} - \frac{0}{4} \right) = \frac{16}{12} = \frac{4}{3} \)Add the results:\( \frac{8}{3} + \frac{4}{3} = \frac{12}{3} = 4 \).
04

Final Step: Solution Conclusion

The final result of the double integral \( \int_{0}^{2} \int_{0}^{x} (x+y^2) \, dy \, dx \) is determined to be 4.

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that studies how things change. It's like a super tool that helps us describe and understand motion and growth.

In this context, we're using calculus to evaluate a double integral, which is essentially the same as finding the area or volume under a surface defined by a function.

Calculus helps us break down complex problems into manageable pieces, making it easier to find precise solutions.
  • Differentiation: This involves finding the rate of change of a function.
  • Integration: This is the process of finding the total accumulation, like area under a curve.
Understanding calculus means grasping how these two fundamental processes interact, and applying them to problems, like finding areas or volumes.
Inner Integral
The inner integral is the first step in solving a double integral problem. It's all about focusing on one variable at a time while treating the other variables as constants.

In the equation \( \int_{0}^{x} (x + y^2) \, dy \), we start by integrating with respect to \( y \). This means we treat \( x \) as a fixed number and find the integral of \( x+y^2 \) over the limits from 0 to \( x \).
  • Separate the terms: \( x \) is treated as a constant during the integration with respect to \( y \).
  • Integrate every term: Calculate separately for \( x \) and \( y^2 \), then combine the results.
  • Apply limits: Substitute in the limits for \( y \), here from 0 to \( x \), to get the expression for just \( x \).
By focusing on one variable, we simplify the problem and set the stage for applying the outer integral.
Outer Integral
Once the inner integral is solved, the result is then used as a new function in the outer integral.

For our exercise, after solving the inner integral, we obtain \( x^2 + \frac{x^3}{3} \). This expression needs to be integrated with respect to \( x \), from 0 to 2.
  • Apply integration rules: Use the simple power rule to integrate each term in the expression.
  • Evaluate between limits: Substitute these limits into the integral result to find the total accumulation.
  • Sum up: Add up the results of each integral to get the final result.
This last step solidifies the solution by finding the total area under the curve from the initial conditions set by the problem.

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