Math Problem Statement

  1. ∫ ∫ π‘₯H𝑒HH𝑑π‘₯ 𝑑� limits of x are from y to 1 and limits of y are from 0 to 1

Solution

We need to evaluate the double integral:

∫01∫y1xex dx dy\int_{0}^{1} \int_{y}^{1} x e^x \, dx \, dy

Step 1: Inner integral

First, evaluate the inner integral:

I=∫y1xex dxI = \int_{y}^{1} x e^x \, dx

For this, use integration by parts. Let:

  • u=xu = x (so du=dxdu = dx)
  • dv=exdxdv = e^x dx (so v=exv = e^x)

Using the integration by parts formula: ∫u dv=uvβˆ’βˆ«v du\int u \, dv = uv - \int v \, du we get: ∫xex dx=xexβˆ’βˆ«ex dx=xexβˆ’ex+C\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C

Thus, the inner integral becomes: I=[xexβˆ’ex]y1I = \left[ x e^x - e^x \right]_{y}^{1}

Substitute the limits of integration (x=yx = y to x=1x = 1): I=(1e1βˆ’e1)βˆ’(yeyβˆ’ey)I = \left( 1 e^1 - e^1 \right) - \left( y e^y - e^y \right) I=(eβˆ’e)βˆ’(yeyβˆ’ey)I = \left( e - e \right) - \left( y e^y - e^y \right) I=0βˆ’(yeyβˆ’ey)I = 0 - (y e^y - e^y) I=eyβˆ’yeyI = e^y - y e^y

Step 2: Outer integral

Now, evaluate the outer integral: ∫01(eyβˆ’yey) dy\int_{0}^{1} \left( e^y - y e^y \right) \, dy

Split this into two separate integrals: ∫01(eyβˆ’yey) dy=∫01ey dyβˆ’βˆ«01yey dy\int_{0}^{1} \left( e^y - y e^y \right) \, dy = \int_{0}^{1} e^y \, dy - \int_{0}^{1} y e^y \, dy

First term:

∫01ey dy=[ey]01=e1βˆ’e0=eβˆ’1\int_{0}^{1} e^y \, dy = \left[ e^y \right]_{0}^{1} = e^1 - e^0 = e - 1

Second term:

Use integration by parts again for ∫01yey dy\int_{0}^{1} y e^y \, dy. Let:

  • u=yu = y (so du=dydu = dy)
  • dv=eydydv = e^y dy (so v=eyv = e^y)

Then: ∫yey dy=yeyβˆ’βˆ«ey dy\int y e^y \, dy = y e^y - \int e^y \, dy ∫yey dy=yeyβˆ’ey+C\int y e^y \, dy = y e^y - e^y + C

Now evaluate from y=0y = 0 to y=1y = 1: ∫01yey dy=[yeyβˆ’ey]01\int_{0}^{1} y e^y \, dy = \left[ y e^y - e^y \right]_{0}^{1} =(1e1βˆ’e1)βˆ’(0e0βˆ’e0)= \left( 1 e^1 - e^1 \right) - \left( 0 e^0 - e^0 \right) =(eβˆ’e)βˆ’(0βˆ’1)= \left( e - e \right) - \left( 0 - 1 \right) =0βˆ’(βˆ’1)=1= 0 - (-1) = 1

Final result:

Substitute the results back into the expression: ∫01(eyβˆ’yey) dy=(eβˆ’1)βˆ’1\int_{0}^{1} \left( e^y - y e^y \right) \, dy = (e - 1) - 1 =eβˆ’2= e - 2

Thus, the value of the double integral is: eβˆ’2\boxed{e - 2}

Would you like further clarification or steps explained in more detail?
Here are 5 related questions to explore:

  1. How does integration by parts work, and why is it useful here?
  2. Can this problem be solved by reversing the order of integration?
  3. What are the geometric implications of this integral?
  4. How do exponentials like exe^x behave under integration?
  5. Can this integral be approximated numerically for different limits?

Tip: For problems involving xexx e^x, remember that integration by parts is the most effective strategy!

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Math Problem Analysis

Mathematical Concepts

Double Integration
Integration by Parts
Exponential Functions

Formulas

Integration by parts formula: ∫ u dv = uv - ∫ v du
Definite integral evaluation: ∫_a^b f(x) dx = F(b) - F(a)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate - Calculus I/II