Math Problem Statement

DAME EL VALOR DE K

El valor de K en la ecuación dada: ∫(de 0 a 1) (10/(2x + 4) - K/(4x + 3)) dx = 5ln(3/2) + (7/4)ln(3/7)

Solution

The equation given in the image is:

01(102x+4K4x+3)dx=5ln(32)+74ln(37)\int_0^1 \left( \frac{10}{2x + 4} - \frac{K}{4x + 3} \right) dx = 5 \ln\left(\frac{3}{2}\right) + \frac{7}{4} \ln\left(\frac{3}{7}\right)

Our goal is to determine the value of KK.

Step-by-Step Solution:

  1. Simplifying the Left-hand Side (Integral):

    01(102x+4K4x+3)dx\int_0^1 \left( \frac{10}{2x + 4} - \frac{K}{4x + 3} \right) dx

    We can simplify the integrand separately: 01102x+4dx01K4x+3dx\int_0^1 \frac{10}{2x + 4} dx - \int_0^1 \frac{K}{4x + 3} dx

    First Integral: For the first term, make the substitution u=2x+4u = 2x + 4, so du=2dxdu = 2dx, and adjust the limits accordingly: x=0u=4,x=1u=6x = 0 \Rightarrow u = 4, \quad x = 1 \Rightarrow u = 6 The integral becomes: 4610u12du=5ln(64)=5ln(32)\int_4^6 \frac{10}{u} \cdot \frac{1}{2} du = 5 \ln\left(\frac{6}{4}\right) = 5 \ln\left(\frac{3}{2}\right)

    Second Integral: For the second term, use the substitution v=4x+3v = 4x + 3, so dv=4dxdv = 4dx, and adjust the limits: x=0v=3,x=1v=7x = 0 \Rightarrow v = 3, \quad x = 1 \Rightarrow v = 7 The integral becomes: 37Kv14dv=K4ln(73)\int_3^7 \frac{K}{v} \cdot \frac{1}{4} dv = \frac{K}{4} \ln\left(\frac{7}{3}\right)

    Therefore, the left-hand side is: 5ln(32)K4ln(73)5 \ln\left(\frac{3}{2}\right) - \frac{K}{4} \ln\left(\frac{7}{3}\right)

  2. Equating the Left-hand Side to the Right-hand Side:

    The equation becomes: 5ln(32)K4ln(73)=5ln(32)+74ln(37)5 \ln\left(\frac{3}{2}\right) - \frac{K}{4} \ln\left(\frac{7}{3}\right) = 5 \ln\left(\frac{3}{2}\right) + \frac{7}{4} \ln\left(\frac{3}{7}\right)

    Canceling 5ln(32)5 \ln\left(\frac{3}{2}\right) from both sides: K4ln(73)=74ln(37)-\frac{K}{4} \ln\left(\frac{7}{3}\right) = \frac{7}{4} \ln\left(\frac{3}{7}\right)

    Since ln(37)=ln(73)\ln\left(\frac{3}{7}\right) = -\ln\left(\frac{7}{3}\right), the equation becomes: K4ln(73)=74ln(73)-\frac{K}{4} \ln\left(\frac{7}{3}\right) = -\frac{7}{4} \ln\left(\frac{7}{3}\right)

    Canceling the common term ln(73)\ln\left(\frac{7}{3}\right) (assuming it's non-zero): K4=74-\frac{K}{4} = -\frac{7}{4}

  3. Solving for KK: K=7K = 7

Final Answer:

The value of KK is 7\boxed{7}.

Do you want more details or have any questions?

Here are five related questions you might want to explore:

  1. How do you handle integrals with more complex rational functions?
  2. Can this method be used for other limits of integration?
  3. What are some practical applications of definite integrals?
  4. How can substitution be generalized for more complex integral problems?
  5. Can logarithmic properties simplify integral evaluations further?

Tip: Always check if the integral limits change after substitution, as it is a common source of errors.

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Math Problem Analysis

Mathematical Concepts

Integration
Definite Integral
Logarithmic Functions
Substitution Method

Formulas

∫(a to b) f(x) dx = F(b) - F(a)
Substitution: u = g(x) => ∫ f(g(x)) g'(x) dx = ∫ f(u) du

Theorems

Fundamental Theorem of Calculus
Properties of Logarithms

Suitable Grade Level

Undergraduate Calculus