Math Problem Statement

Dame el valor de K

Dame el valor de K

Solution

The equation in the image is as follows:

01(45x+2K4x+9)dx=45ln(72)+2ln(913)\int_0^1 \left( \frac{4}{5x+2} - \frac{K}{4x+9} \right) dx = \frac{4}{5} \ln \left( \frac{7}{2} \right) + 2 \ln \left( \frac{9}{13} \right)

To find the value of KK, we need to solve the integral on the left-hand side and equate it to the right-hand side.

Step 1: Evaluate the integral

The integral involves two rational functions:

01(45x+2K4x+9)dx\int_0^1 \left( \frac{4}{5x+2} - \frac{K}{4x+9} \right) dx

We can split this into two integrals:

0145x+2dx01K4x+9dx\int_0^1 \frac{4}{5x+2} dx - \int_0^1 \frac{K}{4x+9} dx

First integral:

0145x+2dx\int_0^1 \frac{4}{5x+2} dx

Let u=5x+2u = 5x + 2, so du=5dxdu = 5 dx, which gives:

45271udu=45ln(72)\frac{4}{5} \int_2^7 \frac{1}{u} du = \frac{4}{5} \ln \left( \frac{7}{2} \right)

Second integral:

01K4x+9dx\int_0^1 \frac{K}{4x+9} dx

Let v=4x+9v = 4x + 9, so dv=4dxdv = 4 dx, which gives:

K49131vdv=K4ln(139)\frac{K}{4} \int_9^{13} \frac{1}{v} dv = \frac{K}{4} \ln \left( \frac{13}{9} \right)

Step 2: Combine the integrals

We now combine both results:

45ln(72)K4ln(139)\frac{4}{5} \ln \left( \frac{7}{2} \right) - \frac{K}{4} \ln \left( \frac{13}{9} \right)

Step 3: Set the equation equal to the right-hand side

Equate this to the right-hand side of the equation:

45ln(72)+2ln(913)\frac{4}{5} \ln \left( \frac{7}{2} \right) + 2 \ln \left( \frac{9}{13} \right)

Step 4: Solve for KK

Simplifying the right-hand side:

2ln(913)=ln(92132)=ln(81169)2 \ln \left( \frac{9}{13} \right) = \ln \left( \frac{9^2}{13^2} \right) = \ln \left( \frac{81}{169} \right)

Thus, the equation becomes:

45ln(72)K4ln(139)=45ln(72)+ln(81169)\frac{4}{5} \ln \left( \frac{7}{2} \right) - \frac{K}{4} \ln \left( \frac{13}{9} \right) = \frac{4}{5} \ln \left( \frac{7}{2} \right) + \ln \left( \frac{81}{169} \right)

Since the terms involving ln(72)\ln \left( \frac{7}{2} \right) cancel out, we are left with:

K4ln(139)=ln(81169)-\frac{K}{4} \ln \left( \frac{13}{9} \right) = \ln \left( \frac{81}{169} \right)

This simplifies to:

K4ln(139)=ln(81169)-\frac{K}{4} \ln \left( \frac{13}{9} \right) = \ln \left( \frac{81}{169} \right)

K4=ln(81169)ln(139)-\frac{K}{4} = \frac{\ln \left( \frac{81}{169} \right)}{\ln \left( \frac{13}{9} \right)}

Since ln(81169)ln(139)=2\frac{\ln \left( \frac{81}{169} \right)}{\ln \left( \frac{13}{9} \right)} = 2, we find that:

K4=2-\frac{K}{4} = 2

Thus,

K=8K = -8

Conclusion:

The value of KK is 8\boxed{-8}.


Would you like more details or have any questions?

Related Questions:

  1. How can you apply substitution techniques in integrals more effectively?
  2. What are some common logarithmic properties that are useful in solving integrals?
  3. How would the answer change if the limits of integration were different?
  4. How can you check if your integral evaluation is correct?
  5. What is the significance of rational functions in calculus problems?

Tip:

When dealing with logarithmic terms inside integrals, always use properties of logarithms to simplify the expressions before attempting to solve them.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Logarithms

Formulas

Integral of rational functions
Logarithmic properties
Substitution in integrals

Theorems

Fundamental Theorem of Calculus
Properties of Logarithms

Suitable Grade Level

Undergraduate Calculus