C. 1-3/5 (6x³+11x² - 4x) 18x² +22x-4) dx

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question

Instructions say to evaluate the integral

### Calculus Integral Problem

Evaluate the following integral:

\[
\int_{0}^{1} \left(6x^3 + 11x^2 - 4x\right)^{-\frac{3}{5}} \left(18x^2 + 22x - 4\right) \, dx 
\]

#### Steps to Solve:
1. **Identify the inner function and its derivative:**
   - Here, let \( u = 6x^3 + 11x^2 - 4x \)
   - Then, \( \frac{du}{dx} = 18x^2 + 22x - 4 \)

2. **Recognize the integral in the form:**
   - The given integrand is of a form that suggests a substitution \( u = 6x^3 + 11x^2 - 4x \), giving \( du = (18x^2 + 22x - 4) dx \)

3. **Transform the limits:**
   - When \( x = 0 \), \( u(0) = 6(0)^3 + 11(0)^2 - 4(0) = 0 \)
   - When \( x = 1 \), \( u(1) = 6(1)^3 + 11(1)^2 - 4(1) = 6 + 11 - 4 = 13 \)

   Thus, the integral becomes:
   \[
   \int_0^{13} u^{-\frac{3}{5}} \, du
   \]

4. **Evaluate the integral:**
   - Use the power rule for integration:
     \[
     \int u^n \, du = \frac{u^{n+1}}{n+1} + C
     \]
   - Here, \( n = -\frac{3}{5} \)
     \[
     \int u^{-\frac{3}{5}} \, du = \frac{u^{-\frac{3}{5} + 1}}{-\frac{3}{5} + 1} = \frac{u^{\frac{2}{5}}}{\frac{2}{5}} = \frac{5}{2} u^{\frac{2}{5}}
     \]

5. **Apply the limits:**
Transcribed Image Text:### Calculus Integral Problem Evaluate the following integral: \[ \int_{0}^{1} \left(6x^3 + 11x^2 - 4x\right)^{-\frac{3}{5}} \left(18x^2 + 22x - 4\right) \, dx \] #### Steps to Solve: 1. **Identify the inner function and its derivative:** - Here, let \( u = 6x^3 + 11x^2 - 4x \) - Then, \( \frac{du}{dx} = 18x^2 + 22x - 4 \) 2. **Recognize the integral in the form:** - The given integrand is of a form that suggests a substitution \( u = 6x^3 + 11x^2 - 4x \), giving \( du = (18x^2 + 22x - 4) dx \) 3. **Transform the limits:** - When \( x = 0 \), \( u(0) = 6(0)^3 + 11(0)^2 - 4(0) = 0 \) - When \( x = 1 \), \( u(1) = 6(1)^3 + 11(1)^2 - 4(1) = 6 + 11 - 4 = 13 \) Thus, the integral becomes: \[ \int_0^{13} u^{-\frac{3}{5}} \, du \] 4. **Evaluate the integral:** - Use the power rule for integration: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \] - Here, \( n = -\frac{3}{5} \) \[ \int u^{-\frac{3}{5}} \, du = \frac{u^{-\frac{3}{5} + 1}}{-\frac{3}{5} + 1} = \frac{u^{\frac{2}{5}}}{\frac{2}{5}} = \frac{5}{2} u^{\frac{2}{5}} \] 5. **Apply the limits:**
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