Math Problem Statement

x2\exponentialE2xdx\int x^2\exponentialE^{2x}dx

∫ x² e^{2x} dx

Solution

To solve the integral x2e2xdx\int x^2 e^{2x} \, dx using integration by parts, let's break down the steps methodically. Integration by parts follows the formula:

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

Step 1: Choose uu and dvdv

We want to simplify the integral by reducing the power of x2x^2, so let's choose:

  • u=x2u = x^2, which implies du=2xdxdu = 2x \, dx
  • dv=e2xdxdv = e^{2x} \, dx, which implies v=12e2xv = \frac{1}{2} e^{2x} (since e2xdx=12e2x\int e^{2x} \, dx = \frac{1}{2} e^{2x})

Step 2: Apply the integration by parts formula

Now apply the formula:

x2e2xdx=x212e2x12e2x2xdx\int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 2x \, dx

Simplifying:

=12x2e2xxe2xdx= \frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx

Step 3: Apply integration by parts again

Now we need to solve xe2xdx\int x e^{2x} \, dx using integration by parts again. Choose:

  • u=xu = x, so du=dxdu = dx
  • dv=e2xdxdv = e^{2x} \, dx, so v=12e2xv = \frac{1}{2} e^{2x}

Apply the formula:

xe2xdx=x12e2x12e2xdx\int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx

=12xe2x14e2x= \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}

Step 4: Substitute back

Now substitute this result into the original equation:

x2e2xdx=12x2e2x(12xe2x14e2x)\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \left( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \right)

Simplify the expression:

x2e2xdx=12x2e2x12xe2x+14e2x+C\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C

Where CC is the constant of integration.

Final Answer:

x2e2xdx=12x2e2x12xe2x+14e2x+C\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C

Do you have any questions or need further clarification on any of the steps?

Here are five questions for further exploration:

  1. How do we determine the best choices for uu and dvdv in integration by parts?
  2. What happens if we choose different values for uu and dvdv in this problem?
  3. Can we verify the solution by differentiating the final result?
  4. What is the difference between using integration by parts once versus multiple times in a problem like this?
  5. How does the constant of integration CC affect the final answer in indefinite integrals?

Tip: When solving integrals involving both polynomials and exponentials, integration by parts is typically an efficient method to reduce the polynomial term step by step.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Exponential Functions

Formulas

∫ u dv = uv - ∫ v du
∫ e^{ax} dx = (1/a) e^{ax}

Theorems

Integration by Parts

Suitable Grade Level

College Level (Calculus II or higher)