Math Problem Statement

find the average value of the function e^y(sqrt4x+e^y), over the rectangle R=[0,3]×[0,2] .

Solution

To find the average value of the function f(x,y)=ey4x+eyf(x, y) = e^y \sqrt{4x + e^y} over the rectangular region R=[0,3]×[0,2]R = [0, 3] \times [0, 2], we use the formula for the average value of a function over a region:

Average value=1Area of RRf(x,y)dA\text{Average value} = \frac{1}{\text{Area of } R} \iint_R f(x, y) \, dA

Step 1: Calculate the Area of RR

The region R=[0,3]×[0,2]R = [0, 3] \times [0, 2] is a rectangle with dimensions 3 and 2, so its area is:

Area of R=3×2=6\text{Area of } R = 3 \times 2 = 6

Step 2: Set Up the Integral

The integral over RR can be expressed as:

Rf(x,y)dA=0203ey4x+eydxdy\iint_R f(x, y) \, dA = \int_0^2 \int_0^3 e^y \sqrt{4x + e^y} \, dx \, dy

Step 3: Evaluate the Inner Integral with Respect to xx

Let's first integrate f(x,y)=ey4x+eyf(x, y) = e^y \sqrt{4x + e^y} with respect to xx from x=0x = 0 to x=3x = 3:

03ey4x+eydx\int_0^3 e^y \sqrt{4x + e^y} \, dx

Using the substitution u=4x+eyu = 4x + e^y, we get du=4dxdu = 4 \, dx, or dx=du4dx = \frac{du}{4}. When x=0x = 0, u=eyu = e^y, and when x=3x = 3, u=43+ey=12+eyu = 4 \cdot 3 + e^y = 12 + e^y. The inner integral becomes:

ey12+eyeyu14du=ey4ey12+eyudu\int_{e^y}^{12 + e^y} e^y \sqrt{u} \cdot \frac{1}{4} \, du = \frac{e^y}{4} \int_{e^y}^{12 + e^y} \sqrt{u} \, du

Step 4: Evaluate the Integral udu\int \sqrt{u} \, du

The antiderivative of u\sqrt{u} is 23u3/2\frac{2}{3} u^{3/2}. Thus,

ey12+eyudu=[23u3/2]ey12+ey=23((12+ey)3/2(ey)3/2)\int_{e^y}^{12 + e^y} \sqrt{u} \, du = \left[ \frac{2}{3} u^{3/2} \right]_{e^y}^{12 + e^y} = \frac{2}{3} \left( (12 + e^y)^{3/2} - (e^y)^{3/2} \right)

Step 5: Substitute Back and Complete the Integral

Substituting back, the inner integral becomes:

ey423((12+ey)3/2(ey)3/2)=ey6((12+ey)3/2(ey)3/2)\frac{e^y}{4} \cdot \frac{2}{3} \left( (12 + e^y)^{3/2} - (e^y)^{3/2} \right) = \frac{e^y}{6} \left( (12 + e^y)^{3/2} - (e^y)^{3/2} \right)

Now, the average value of f(x,y)f(x, y) over RR is:

1602ey6((12+ey)3/2(ey)3/2)dy\frac{1}{6} \int_0^2 \frac{e^y}{6} \left( (12 + e^y)^{3/2} - (e^y)^{3/2} \right) \, dy

This expression represents the average value of f(x,y)f(x, y) over RR. Calculating it by hand is challenging, so numerical integration would provide an exact answer.

Would you like further clarification on the steps or a numeric approximation?


  1. How would the calculation change if RR had different bounds?
  2. What substitution techniques are useful for functions similar to ey4x+eye^y \sqrt{4x + e^y}?
  3. How does changing the exponent in the square root impact integration?
  4. What is the impact of the eye^y factor on the average value over larger regions?
  5. What happens if we add constraints to xx or yy in region RR?

Tip: Substitution is a useful tool for simplifying square root and exponential terms in integrals.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Integration
Average Value of a Function

Formulas

Average value of a function over region R: (1/Area of R) ∫∫_R f(x, y) dA
Area of a rectangle: width * height

Theorems

Fubini's Theorem

Suitable Grade Level

Undergraduate