Math Problem Statement

Evaluate the integral \( \int_{1}^{e^4} \frac{\ln x}{x} \, dx \).

Solution

The integral presented is:

1e4lnxxdx\int_{1}^{e^4} \frac{\ln x}{x} \, dx

This type of integral is related to the differentiation of logarithmic functions. Let's solve it step by step.

Solution:

  1. Substitution: Let u=lnxu = \ln x. Then: dudx=1x    dx=xdu.\frac{du}{dx} = \frac{1}{x} \implies dx = x \, du. Also, when x=1x = 1, u=ln1=0u = \ln 1 = 0, and when x=e4x = e^4, u=ln(e4)=4u = \ln(e^4) = 4.

  2. Rewrite the integral: Substituting u=lnxu = \ln x and dx=xdudx = x \, du: 1e4lnxxdx=04udu.\int_{1}^{e^4} \frac{\ln x}{x} \, dx = \int_{0}^{4} u \, du.

  3. Evaluate the integral: The integral of uu is: udu=u22.\int u \, du = \frac{u^2}{2}. Apply the limits u=0u = 0 and u=4u = 4: u2204=422022=1620=8.\frac{u^2}{2} \bigg|_{0}^{4} = \frac{4^2}{2} - \frac{0^2}{2} = \frac{16}{2} - 0 = 8.

Final Answer:

1e4lnxxdx=8.\int_{1}^{e^4} \frac{\ln x}{x} \, dx = 8.

Would you like a detailed explanation of any specific step or more examples related to this problem?


Here are 5 related questions for deeper understanding:

  1. How does substitution simplify solving integrals involving logarithms?
  2. Can you solve lnxx2dx\int \frac{\ln x}{x^2} \, dx using substitution?
  3. How would the answer change if the upper limit was e5e^5 instead of e4e^4?
  4. What other functions are commonly integrated using substitution?
  5. Can this integral be solved using integration by parts? Why or why not?

Tip: When encountering integrals involving lnx\ln x, look for patterns involving its derivative 1x\frac{1}{x}, as substitution often simplifies these problems significantly.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Logarithmic Functions
Substitution Method

Formulas

\( \int \frac{\ln x}{x} \, dx = \frac{(\ln x)^2}{2} + C \)
Substitution: \( u = \ln x \), \( du = \frac{1}{x} dx \)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or early college