Math Problem Statement

Find the average value of f over the given rectangle: 1. f(x, y) = 5e^y sqrt(x + e^y), R = [0, 6] x [0, 1] 2. f(x, y) = 3e^y sqrt(x + e^y), R = [0, 10] x [0, 1]

Solution

Thank you for sharing the image. From the screenshot, I see two problems requiring the calculation of the average value of a function over a given rectangle using double integrals. Let's tackle one problem at a time.


Problem 15:

The function and domain are:

  • f(x,y)=5eyx+eyf(x, y) = 5e^y \sqrt{x + e^y}
  • The rectangle R=[0,6]×[0,1]R = [0, 6] \times [0, 1].

The average value of ff over RR is given by: favg=1Area of RRf(x,y)dAf_{\text{avg}} = \frac{1}{\text{Area of } R} \int \int_R f(x, y) \, dA Here, the area of RR is: Area of R=(60)(10)=6\text{Area of } R = (6 - 0)(1 - 0) = 6

Substitute into the formula: favg=1601065eyx+eydxdyf_{\text{avg}} = \frac{1}{6} \int_0^1 \int_0^6 5e^y \sqrt{x + e^y} \, dx \, dy

Step 1: Integrate with respect to xx

Inside the integral, treat yy as constant. The inner integral becomes: 065eyx+eydx\int_0^6 5e^y \sqrt{x + e^y} \, dx Using substitution, let u=x+eyu = x + e^y, so du=dxdu = dx and the limits become:

  • When x=0x = 0: u=eyu = e^y,
  • When x=6x = 6: u=6+eyu = 6 + e^y.

The integral becomes: 065eyx+eydx=ey6+ey5eyudu\int_0^6 5e^y \sqrt{x + e^y} \, dx = \int_{e^y}^{6 + e^y} 5e^y \sqrt{u} \, du

Now calculate: 5eyudu=5ey23u3/2=10ey3u3/2\int 5e^y \sqrt{u} \, du = 5e^y \cdot \frac{2}{3} u^{3/2} = \frac{10e^y}{3} u^{3/2}

Evaluate from u=eyu = e^y to u=6+eyu = 6 + e^y: 10ey3[(6+ey)3/2(ey)3/2]\frac{10e^y}{3} \left[ (6 + e^y)^{3/2} - (e^y)^{3/2} \right]

Step 2: Integrate with respect to yy

Substitute this result into the outer integral: favg=160110ey3[(6+ey)3/2(ey)3/2]dyf_{\text{avg}} = \frac{1}{6} \int_0^1 \frac{10e^y}{3} \left[ (6 + e^y)^{3/2} - (e^y)^{3/2} \right] \, dy Simplify: favg=5901ey[(6+ey)3/2(ey)3/2]dyf_{\text{avg}} = \frac{5}{9} \int_0^1 e^y \left[ (6 + e^y)^{3/2} - (e^y)^{3/2} \right] \, dy

This integral will require numerical methods or further substitution for evaluation.


Problem 16:

The function and domain are:

  • f(x,y)=3eyx+eyf(x, y) = 3e^y \sqrt{x + e^y}
  • The rectangle R=[0,10]×[0,1]R = [0, 10] \times [0, 1].

Following the same method as above:

  1. The area of RR is (100)(10)=10(10 - 0)(1 - 0) = 10.
  2. The average value formula becomes: favg=110010103eyx+eydxdyf_{\text{avg}} = \frac{1}{10} \int_0^1 \int_0^{10} 3e^y \sqrt{x + e^y} \, dx \, dy

The inner integral is: 0103eyx+eydx\int_0^{10} 3e^y \sqrt{x + e^y} \, dx Using the same substitution u=x+eyu = x + e^y, calculate and proceed similarly to Problem 15.

Would you like me to simplify the computations further for any of these problems? Let me know!


Related Questions:

  1. How do you set up double integrals for average value calculations?
  2. What is the substitution method in integrals, and how does it simplify problems like these?
  3. Can numerical integration techniques help solve complex integrals more efficiently?
  4. What is the importance of understanding the bounds of integration in these problems?
  5. How can software tools like MATLAB or Python be used to evaluate such integrals?

Tip: Always break down double integrals into simpler single integrals by carefully choosing the order of integration!

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Math Problem Analysis

Mathematical Concepts

Double Integrals
Average Value of Functions
Substitution in Integrals

Formulas

f_avg = (1/Area of R) ∫∫_R f(x, y) dA
Area of R = (width) x (height)

Theorems

Fundamental Theorem of Calculus for Double Integrals

Suitable Grade Level

Grades 11-12, College-level Calculus