Math Problem Statement

∫π‘₯ 3 𝑒2π‘₯𝑑π‘₯

Solution

The integral you provided is:

∫x3e2x dx\int x^3 e^{2x} \, dx

To solve this, we use integration by parts multiple times. Integration by parts is based on the formula:

∫u dv=uvβˆ’βˆ«v du\int u \, dv = uv - \int v \, du

Step 1: Choose uu and dvdv

Let:

  • u=x3u = x^3, so du=3x2 dxdu = 3x^2 \, dx
  • dv=e2x dxdv = e^{2x} \, dx, so v=12e2xv = \frac{1}{2}e^{2x} (since the derivative of e2xe^{2x} is 2e2x2e^{2x})

Step 2: Apply Integration by Parts

Now apply the integration by parts formula:

∫x3e2x dx=12x3e2xβˆ’βˆ«123x2e2x dx\int x^3 e^{2x} \, dx = \frac{1}{2} x^3 e^{2x} - \int \frac{1}{2} 3x^2 e^{2x} \, dx

Simplify:

=12x3e2xβˆ’32∫x2e2x dx= \frac{1}{2} x^3 e^{2x} - \frac{3}{2} \int x^2 e^{2x} \, dx

Step 3: Repeat the process for ∫x2e2x dx\int x^2 e^{2x} \, dx

Let:

  • u=x2u = x^2, so du=2x dxdu = 2x \, dx
  • dv=e2x dxdv = e^{2x} \, dx, so v=12e2xv = \frac{1}{2}e^{2x}

Now apply integration by parts again:

∫x2e2x dx=12x2e2xβˆ’βˆ«122xe2x dx\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \int \frac{1}{2} 2x e^{2x} \, dx

Simplify:

=12x2e2xβˆ’12∫2xe2x dx= \frac{1}{2} x^2 e^{2x} - \frac{1}{2} \int 2x e^{2x} \, dx

Step 4: Repeat for ∫xe2x dx\int x e^{2x} \, dx

Let:

  • u=xu = x, so du=dxdu = dx
  • dv=e2x dxdv = e^{2x} \, dx, so v=12e2xv = \frac{1}{2}e^{2x}

Now apply integration by parts:

∫xe2x dx=12xe2xβˆ’12∫e2x dx\int x e^{2x} \, dx = \frac{1}{2} x e^{2x} - \frac{1}{2} \int e^{2x} \, dx

The last integral is straightforward:

∫e2x dx=12e2x\int e^{2x} \, dx = \frac{1}{2}e^{2x}

So:

∫xe2x dx=12xe2xβˆ’14e2x\int x e^{2x} \, dx = \frac{1}{2} x e^{2x} - \frac{1}{4}e^{2x}

Step 5: Combine all results

Now substitute back into the original equation:

∫x2e2x dx=12x2e2xβˆ’12(12xe2xβˆ’14e2x)\int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} \left( \frac{1}{2} x e^{2x} - \frac{1}{4}e^{2x} \right)

Simplify:

=12x2e2xβˆ’14xe2x+18e2x= \frac{1}{2} x^2 e^{2x} - \frac{1}{4} x e^{2x} + \frac{1}{8}e^{2x}

Finally, substitute this back into the previous result:

∫x3e2x dx=12x3e2xβˆ’32(12x2e2xβˆ’14xe2x+18e2x)\int x^3 e^{2x} \, dx = \frac{1}{2} x^3 e^{2x} - \frac{3}{2} \left( \frac{1}{2} x^2 e^{2x} - \frac{1}{4} x e^{2x} + \frac{1}{8}e^{2x} \right)

Simplify:

∫x3e2x dx=12x3e2xβˆ’34x2e2x+38xe2xβˆ’316e2x+C\int x^3 e^{2x} \, dx = \frac{1}{2} x^3 e^{2x} - \frac{3}{4} x^2 e^{2x} + \frac{3}{8} x e^{2x} - \frac{3}{16}e^{2x} + C

Final Answer:

∫x3e2x dx=12x3e2xβˆ’34x2e2x+38xe2xβˆ’316e2x+C\int x^3 e^{2x} \, dx = \frac{1}{2} x^3 e^{2x} - \frac{3}{4} x^2 e^{2x} + \frac{3}{8} x e^{2x} - \frac{3}{16}e^{2x} + C

Would you like any further details or clarification on any step?


Here are 5 related questions:

  1. How do you choose the uu and dvdv in integration by parts?
  2. Can integration by parts be used for other transcendental functions like sin⁑(x)\sin(x)?
  3. What is the general formula for integration by parts?
  4. How does repeated integration by parts work in practical applications?
  5. What are the conditions under which integration by parts simplifies a problem?

Tip: When integrating by parts multiple times, always check for a recursive pattern to simplify calculations.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Exponential Functions

Formulas

Integration by Parts Formula: ∫ u dv = uv - ∫ v du
Exponential Function: e^{2x}
General Integration Techniques

Theorems

Integration by Parts Theorem

Suitable Grade Level

University Level (Calculus II or III)