Math Problem Statement

Evaluate the definite integral \( \int_{4}^{5} (x - 2)e^x \, dx \).

Solution

To solve the given integral, which is:

45(x2)exdx\int_{4}^{5} (x - 2)e^x \, dx

we'll use integration by parts, which states:

udv=uvvdu\int u \, dv = uv - \int v \, du

Step-by-step solution:

  1. Choose uu and dvdv: Let: u=(x2),dv=exdxu = (x - 2), \quad dv = e^x \, dx

  2. Compute dudu and vv: du=dx,v=exdx=exdu = dx, \quad v = \int e^x \, dx = e^x

  3. Apply the integration by parts formula: Substituting into the formula: (x2)exdx=(x2)exexdx\int (x - 2)e^x \, dx = (x - 2)e^x - \int e^x \, dx

  4. Simplify further: The second term simplifies as follows: exdx=ex\int e^x \, dx = e^x Thus: (x2)exdx=(x2)exex\int (x - 2)e^x \, dx = (x - 2)e^x - e^x

  5. Combine terms: (x2)exdx=(x3)ex\int (x - 2)e^x \, dx = (x - 3)e^x

  6. Evaluate the definite integral: Now evaluate from x=4x = 4 to x=5x = 5: 45(x2)exdx=[(x3)ex]45\int_{4}^{5} (x - 2)e^x \, dx = \left[(x - 3)e^x \right]_{4}^{5}

    Compute the value at x=5x = 5 and x=4x = 4:

    • At x=5x = 5: (53)e5=2e5(5 - 3)e^5 = 2e^5
    • At x=4x = 4: (43)e4=1e4=e4(4 - 3)e^4 = 1e^4 = e^4

    So: 45(x2)exdx=2e5e4\int_{4}^{5} (x - 2)e^x \, dx = 2e^5 - e^4

Final Answer:

2e5e4\boxed{2e^5 - e^4}


Would you like further clarification or numerical approximation?

Here are 5 related questions:

  1. What is the principle behind integration by parts, and how is it applied?
  2. Can integration by parts be applied iteratively for more complex functions?
  3. How do we handle definite integrals with exponential functions?
  4. What happens if x2x - 2 is replaced with a polynomial of higher degree?
  5. What is the approximate value of 2e5e42e^5 - e^4 using e2.718e \approx 2.718?

Tip: Always simplify the integrand before applying integration by parts to reduce complexity.

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Math Problem Analysis

Mathematical Concepts

Integration
Integration by Parts
Exponential Functions

Formulas

Integration by parts: \( \int u \, dv = uv - \int v \, du \)

Theorems

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Suitable Grade Level

Grades 11-12