Chapter 13: Problem 5
Evaluate the integrals. a. \(\int_{0}^{1} \int_{2}^{4} x y^{2} d y d x\) b. \(\int_{2}^{4} \int_{0}^{1} x y^{2} d x d y\) c. \(\quad \int_{0}^{1} \int_{2}^{4} x y^{2} d x d y\) d. \(\int_{0}^{1} \int_{0}^{y} x y^{2} d x d y\) e. \(\int_{0}^{1} \int_{x^{2}}^{x} x y d y d x\) f. \(\int_{1}^{4} \int_{y}^{y^{2}} x^{2}+y^{2} d x d y\) g. \(\quad \int_{1}^{2} \int_{e^{-x}}^{e^{x}} \frac{x}{y} d y d x\) h. \(\int_{0}^{\sqrt{3}} \int_{1}^{4-x^{2}} x+y d y d x\) i. \(\int_{0}^{1} \int_{1-x^{2}}^{4-x^{2}} x+y d y d x\)
Short Answer
Step by step solution
Solve Integral a
Solve Integral b
Solve Integral c
Solve Integral d
Solve Integral e
Solve Integral f
Solve Integral g
Solve Integral h
Solve Integral i
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integrals
- Double integrals extend the concept of integration from one-dimensional to two-dimensional areas.
- They are used for computing the area of a surface or total values over specific regions.
- In practice, one computes iterated integrals where the integral is done one variable at a time.
Iterated Integrals
- Identifying the limits for the inner integral, which is evaluated first.
- Computing the result of this inner integral, treating other variables as constants.
- Using the result of the inner integral as the integrand for the outer integral.
Integration Limits
- The limits for each integral (inner and outer) indicate where integration starts and ends for the corresponding variable.
- Inner limits often depend on the outer variable, reflecting region boundaries that are variable-specific, like curves or lines on the xy-plane.
- Outer limits are typically constants or defined by intersections of the boundary curves and lines.
Definite Integration
- Definite integrals yield a concrete numerical result, often representing physical quantities like area, volume, or total mass.
- When dealing with definite integrals, it's crucial to evaluate the antiderivative at the upper limit and subtract its value at the lower limit.
- In double integrals, this same process is applied iteratively for each dimension, leading to a solution that encompasses the designated two-dimensional region or surface.