Math Problem Statement

Evaluate. Check by differentiating.

Integral from nothing to nothing 9 x e Superscript 5 x Baseline dx

Solution

The integral given is:

9xe5xdx\int 9x e^{5x} \, dx

Step 1: Solve the integral

To evaluate this integral, we will use integration by parts. The formula for integration by parts is:

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

Here, we let:

  • u=9xu = 9x (so that du=9dxdu = 9 \, dx),
  • dv=e5xdxdv = e^{5x} \, dx (so that v=15e5xv = \frac{1}{5} e^{5x}).

Using the integration by parts formula:

9xe5xdx=9x15e5x15e5x9dx\int 9x e^{5x} \, dx = 9x \cdot \frac{1}{5} e^{5x} - \int \frac{1}{5} e^{5x} \cdot 9 \, dx

Simplify the terms:

9xe5xdx=9x5e5x95e5xdx\int 9x e^{5x} \, dx = \frac{9x}{5} e^{5x} - \frac{9}{5} \int e^{5x} \, dx

The integral of e5xe^{5x} is 15e5x\frac{1}{5} e^{5x}, so:

9xe5xdx=9x5e5x9515e5x\int 9x e^{5x} \, dx = \frac{9x}{5} e^{5x} - \frac{9}{5} \cdot \frac{1}{5} e^{5x}

Simplify further:

9xe5xdx=9x5e5x925e5x\int 9x e^{5x} \, dx = \frac{9x}{5} e^{5x} - \frac{9}{25} e^{5x}

Combine terms:

9xe5xdx=e5x(9x5925)\int 9x e^{5x} \, dx = e^{5x} \left( \frac{9x}{5} - \frac{9}{25} \right)

This is the indefinite integral. Since the integral is from 0 to 0, the evaluation yields 0 because both the upper and lower bounds are the same.

Step 2: Check by Differentiating

To confirm the indefinite integral, differentiate the result:

ddx(e5x(9x5925))\frac{d}{dx} \left( e^{5x} \left( \frac{9x}{5} - \frac{9}{25} \right) \right)

Using the product rule, let:

  • First term: u=e5xu = e^{5x},
  • Second term: v=9x5925v = \frac{9x}{5} - \frac{9}{25}.

Differentiate: ddx(e5x)=5e5x,ddx(9x5925)=95.\frac{d}{dx} \left( e^{5x} \right) = 5e^{5x}, \quad \frac{d}{dx} \left( \frac{9x}{5} - \frac{9}{25} \right) = \frac{9}{5}.

Apply the product rule:

ddx(e5x(9x5925))=e5x95+5e5x(9x5925)\frac{d}{dx} \left( e^{5x} \left( \frac{9x}{5} - \frac{9}{25} \right) \right) = e^{5x} \cdot \frac{9}{5} + 5e^{5x} \cdot \left( \frac{9x}{5} - \frac{9}{25} \right)

Simplify:

ddx=e5x95+e5x(9x95)\frac{d}{dx} = e^{5x} \cdot \frac{9}{5} + e^{5x} \cdot \left( 9x - \frac{9}{5} \right)

Factor e5xe^{5x}:

ddx=e5x(95+9x95)\frac{d}{dx} = e^{5x} \left( \frac{9}{5} + 9x - \frac{9}{5} \right)

Combine terms:

ddx=e5x9x\frac{d}{dx} = e^{5x} \cdot 9x

This matches the original function 9xe5x9x e^{5x}. Hence, the integral is correct.


Final Answer

009xe5xdx=0\int_{0}^{0} 9x e^{5x} \, dx = 0

Do you have any questions or need further explanation? 😊

Here are 5 follow-up questions:

  1. Can you explain integration by parts in more detail?
  2. How do we determine uu and dvdv for integration by parts?
  3. What is the general formula for integrating xeaxx e^{ax}?
  4. How can we apply the limits more directly instead of integrating fully?
  5. What are other methods for solving such integrals?

Tip: Always verify integrals by differentiating to ensure accuracy!

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Math Problem Analysis

Mathematical Concepts

Integration
Differentiation
Integration by Parts
Exponential Functions

Formulas

Integration by Parts: ∫u dv = uv - ∫v du
Exponential Function Integral: ∫e^(ax) dx = (1/a)e^(ax) + C
Derivative of a Product: d(uv)/dx = u dv/dx + v du/dx

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or early college