Math Problem Statement

Compute the value of the integral: ∫ from 1 to 4 of 3 ln(9 + 3s) ds.

Solution

We are tasked with evaluating the following definite integral:

143ln(9+3s)ds\int_1^4 3 \ln(9 + 3s) \, ds

Let's solve this step by step:

Step 1: Substitution

To simplify the integral, we use substitution. Let: u=9+3su = 9 + 3s Thus, the derivative of uu with respect to ss is: du=3dsords=du3du = 3 \, ds \quad \text{or} \quad ds = \frac{du}{3}

Also, update the limits of integration according to the substitution:

  • When s=1s = 1, u=9+3(1)=12u = 9 + 3(1) = 12,
  • When s=4s = 4, u=9+3(4)=21u = 9 + 3(4) = 21.

Step 2: Update the integral

Substitute into the integral: 143ln(9+3s)ds=12213ln(u)du3\int_{1}^{4} 3 \ln(9 + 3s) \, ds = \int_{12}^{21} 3 \ln(u) \cdot \frac{du}{3}

The 33's cancel out, so we are left with: 1221ln(u)du\int_{12}^{21} \ln(u) \, du

Step 3: Integration by parts

We need to integrate ln(u)\ln(u). This is a standard integral that can be solved using integration by parts. Let: v=ln(u)anddw=duv = \ln(u) \quad \text{and} \quad dw = du Then, dv=1uduandw=udv = \frac{1}{u} \, du \quad \text{and} \quad w = u

Now, apply the integration by parts formula: vdw=vwwdv\int v \, dw = v \cdot w - \int w \, dv Substituting the terms: ln(u)du=uln(u)u1udu=uln(u)u+C\int \ln(u) \, du = u \ln(u) - \int u \cdot \frac{1}{u} \, du = u \ln(u) - u + C

Step 4: Evaluate the integral

Now, evaluate this at the limits u=21u = 21 and u=12u = 12: [uln(u)u]1221=(21ln(21)21)(12ln(12)12)\left[ u \ln(u) - u \right]_{12}^{21} = \left( 21 \ln(21) - 21 \right) - \left( 12 \ln(12) - 12 \right)

Step 5: Approximate the result

Let's compute the numerical values of this expression.

21ln(21)21×3.0445=63.934521 \ln(21) \approx 21 \times 3.0445 = 63.9345 12ln(12)12×2.4849=29.818812 \ln(12) \approx 12 \times 2.4849 = 29.8188 Now substitute: (63.934521)(29.818812)=42.934517.8188=25.1157(63.9345 - 21) - (29.8188 - 12) = 42.9345 - 17.8188 = 25.1157

Thus, the value of the integral is approximately: 25.12\boxed{25.12}

Would you like further clarification or a more detailed breakdown of any step?

Here are 5 related questions for further practice:

  1. How would the integral change if the bounds were from 0 to 3 instead of 1 to 4?
  2. Can you compute the integral ln(3x+5)dx\int \ln(3x + 5) \, dx?
  3. How does the substitution method simplify complex integrals?
  4. What are the main steps of integration by parts, and when should it be used?
  5. What if the function inside the logarithm was quadratic instead of linear? How would that affect the solution?

Tip: Always check if substitution can simplify the integration before using more advanced techniques like integration by parts.

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Math Problem Analysis

Mathematical Concepts

Definite Integration
Substitution Method
Integration by Parts
Natural Logarithms

Formulas

∫ ln(x) dx = x ln(x) - x + C
Substitution: u = 9 + 3s

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate (Calculus I/II)