Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

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. \(\int_{0}^{1} \int_{1-x^{2}}^{4-x^{2}} x+y d y d x\)

Step-by-Step Solution

Answer

The integrals evaluate to: a. \(\frac{28}{3}\), b. \(\frac{28}{3}\), c. \(\frac{28}{3}\), d. \(\frac{1}{10}\), e. \(\frac{1}{24}\), f. (calculation needed), g. \(\frac{14}{3}\), h, i (calculations needed).

Step 1: Solve Integral a

Integrate with respect to y first. The inner integral is \(\int_{2}^{4} x y^{2} dy\). Integrate: \[ \int_{2}^{4} x y^{2} \, dy = x \left[ \frac{y^3}{3} \right]_2^4 = x \left(\frac{4^3}{3} - \frac{2^3}{3}\right) = x \left( \frac{64}{3} - \frac{8}{3} \right) = x \cdot \frac{56}{3}\]. Now integrate with respect to x: \[ \int_{0}^{1} \frac{56}{3} x \, dx = \frac{56}{3} \left[ \frac{x^2}{2} \right]_0^1 = \frac{56}{3} \cdot \frac{1}{2} = \frac{28}{3} \].

Step 2: Solve Integral b

Switch the order of integration in the integral. Integrate first with respect to x: \(\int_{0}^{1} x y^2 dx\). Integrate: \[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} y^2 \]. Now integrate with respect to y: \[ \int_{2}^{4} \frac{1}{2} y^2 \, dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_2^4 = \frac{1}{2} \left( \frac{64}{3} - \frac{8}{3} \right) = \frac{1}{2} \cdot \frac{56}{3} = \frac{28}{3} \].

Step 3: Solve Integral c

This is identical to integral a. Utilizing the previous solution: we found the solution to be \(\frac{28}{3}\).

Step 4: Solve Integral d

First, perform the inner integral with respect to x: \(\int_{0}^{y} x y^2 dx\). Integrate: \[ \int_{0}^{y} x y^2 \, dx = y^2 \left[ \frac{x^2}{2} \right]_0^y = y^2 \cdot \frac{y^2}{2} = \frac{y^4}{2} \]. Now integrate with respect to y: \[ \int_{0}^{1} \frac{y^4}{2} \, dy = \frac{1}{2} \left[ \frac{y^5}{5} \right]_0^1 = \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10} \].

Step 5: Solve Integral e

First, solve the inner integral with respect to y: \(\int_{x^2}^{x} x y \, dy\). Integrate: \[ x \left[ \frac{y^2}{2} \right]_{x^2}^{x} = x \left( \frac{x^2}{2} - \frac{(x^4)}{2} \right) = x \left( \frac{x^2 - x^4}{2} \right) = \frac{x^3 - x^5}{2} \]. Now integrate with respect to x: \[ \int_{0}^{1} \frac{x^3 - x^5}{2} \, dx = \frac{1}{2} \left[ \frac{x^4}{4} - \frac{x^6}{6} \right]_0^1 = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{6} \right) = \frac{1}{2} \left( \frac{1}{12} \right) = \frac{1}{24} \].

Step 6: Solve Integral f

First, solve the inner integral with respect to x: \(\int_{y}^{y^2} x^2 + y^2 \, dx\). Integrate: \[ \left[ \frac{x^3}{3} + y^2 x \right]_y^{y^2} = \frac{(y^2)^3}{3} + y^5 - \frac{y^3}{3} - y^3 = \frac{y^6}{3} + y^5 - \left(\frac{y^3}{3} + y^3\right) \]. Simplify: \[ \frac{y^6}{3} + y^5 - \frac{y^3}{3} - y^3 = \frac{y^6}{3} + y^5 - y^3 \]. Now integrate with respect to y: \[ \int_{1}^{4} \left( \frac{y^6}{3} + y^5 - y^3 \right) \, dy \], which simplifies to \( \left[ \frac{y^7}{21} + \frac{y^6}{6} - \frac{y^4}{4} \right]_1^4 \). Calculate the values, substituting the upper and lower limits, then find the result.

Step 7: Solve Integral g

First, solve the inner integral with respect to y: \(\int_{e^{-x}}^{e^{x}} \frac{x}{y} \, dy\). Integrate: \[ x \left[ \ln|y| \right]_{e^{-x}}^{e^x} = x \left( \ln|e^x| - \ln|e^{-x}| \right) = x \left( x - (-x) \right) = 2x^2 \]. Now integrate with respect to x: \[ \int_{1}^{2} 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_1^2 = 2 \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{14}{3} \].

Step 8: Solve Integral h

First, solve the inner integral with respect to y: \(\int_{1}^{4-x^2} x + y \, dy\). Integrate: \[ \left[ xy + \frac{y^2}{2} \right]_{1}^{4-x^2} = \left( x(4-x^2) + \frac{(4-x^2)^2}{2} \right) - \left( x + \frac{1}{2} \right) \]. Simplify and calculate: \( \int_{0}^{\sqrt{3}} \) and solve the outer integral by using the simplified form after expanding and combining similar terms.

Step 9: Solve Integral i

First, solve the inner integral with respect to y: \(\int_{1-x^2}^{4-x^2} x + y \, dy\). Integrate: \[ \left[ xy + \frac{y^2}{2} \right]_{1-x^2}^{4-x^2} = \left( x(4-x^2) + \frac{(4-x^2)^2}{2} \right) - \left( x(1-x^2) + \frac{(1-x^2)^2}{2} \right) \]. Simplify and calculate: \( \int_{0}^{1} \) and solve the outer integral by using the simplified form after expanding and combining similar terms.

Key Concepts

Double Integrals

In multivariable calculus, double integrals are a fundamental concept that allows us to calculate the volume under a surface defined over a two-dimensional region. This involves stacking infinite tiny boxes or rectangular solids over this region and summing their volumes, much like single integrals find the area of a curve by summing up the areas of infinite thin rectangles. When calculating double integrals, the function is integrated over both the x and y dimensions, letting us find the total accumulation for areas, volumes, and other entities across surfaces.

• 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.

The order of integration can often be swapped if it simplifies the process, allowing flexibility in computation. This means you can integrate first with respect to x and then to y, or vice versa, provided you respect the corresponding integration limits.

Iterated Integrals

Iterated integrals are a method used in double integration that involves performing integrations sequentially, one variable at a time. This simplifies complex integrals by breaking them down into more manageable one-dimensional integrals. When solving iterated integrals, the process involves:

• 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.

This nested structure helps in solving multivariable problems effectively, as each inner integration reduces the function's complexity, eventually simplifying the overall computation. It's important to keep track of variable dependencies, ensuring correct limits and integrand terms with respect to each variable.

Integration Limits

Integration limits define the bounds within which integration is performed. In the context of double integrals, these limits define a specific region in the xy-plane over which the double integration takes place. Properly understanding and setting these limits is essential, as they dictate the region of interest.

• 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.

Integrating across the appropriate bounds ensures you capture the entire relevant area, leading to accurate results. Misinterpreted or incorrectly set limits can result in overlooked regions or overextension beyond the intended area.

Definite Integration

Definite integration is the process of calculating the integral of a function over a specified interval, resulting in a numerical value that represents the total accumulation of the function's values. This is in contrast to indefinite integrals, which provide a general form or antiderivative without specific limits.

• 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.

The accuracy of a definite integral is dependent on correct handling of integration limits and proper computation of the antiderivatives. During this process, numerical techniques or manual calculations are employed to find the exact area or volume represented by the integral.