Math Problem Statement

Evaluate the indefinite integral: \(\int (1 - x) e^x \, dx\).

Solution

The image you uploaded contains a mathematical problem involving the calculation of an integral. The problem is asking you to evaluate the following indefinite integral:

(1x)exdx\int \left(1 - x\right) e^{x} \, dx

Solution:

To solve the integral (1x)exdx\int \left(1 - x\right) e^{x} \, dx, we can use integration by parts. The formula for integration by parts is:

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

Step 1: Identify uu and dvdv

Let:

  • u=1xu = 1 - x, hence du=dxdu = -dx
  • dv=exdxdv = e^x \, dx, hence v=exv = e^x

Step 2: Apply the integration by parts formula

(1x)exdx=(1x)exex(dx)\int \left(1 - x\right) e^{x} \, dx = (1 - x)e^x - \int e^x(-dx)

Step 3: Simplify the expression

(1x)exdx=(1x)ex+exdx\int \left(1 - x\right) e^{x} \, dx = (1 - x)e^x + \int e^x \, dx

exdx=ex\int e^x \, dx = e^x

Thus:

(1x)exdx=(1x)ex+ex+C\int \left(1 - x\right) e^{x} \, dx = (1 - x)e^x + e^x + C

(1x)exdx=ex(1x+1)+C\int \left(1 - x\right) e^{x} \, dx = e^x(1 - x + 1) + C

(1x)exdx=ex(2x)+C\int \left(1 - x\right) e^{x} \, dx = e^x(2 - x) + C

Where CC is the constant of integration.

Final Answer:

(1x)exdx=ex(2x)+C\int \left(1 - x\right) e^{x} \, dx = e^x(2 - x) + C

Would you like further details on any of these steps?

Relative Questions:

  1. How do you apply integration by parts to more complex integrals?
  2. What is the method for solving integrals involving products of polynomials and exponentials?
  3. Can you explain the importance of the constant of integration in indefinite integrals?
  4. How does the choice of uu and dvdv affect the integration by parts process?
  5. What are some common mistakes to avoid when performing integration by parts?

Tip: Always double-check your uu and dvdv choices in integration by parts to ensure the integration simplifies the problem effectively.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Integration by Parts

Formulas

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

Theorems

-

Suitable Grade Level

Grades 11-12