Integration by Parts: Calculus Integral Help?

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SUMMARY

The integral of the function \( x^2 \ln(x) \) from 1 to 2 is evaluated using integration by parts. By setting \( u = \ln(x) \) and \( dv = x^2 dx \), the integral simplifies to \( \int_1^2 x^2 \ln(x) \, dx = \frac{8}{3} \ln(2) - \frac{7}{9} \). The final result is \( \frac{8}{3} \ln(2) - \frac{7}{9} \), demonstrating the application of integration by parts effectively.

PREREQUISITES
  • Understanding of integration by parts
  • Familiarity with natural logarithms
  • Basic knowledge of definite integrals
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the method of integration by parts in detail
  • Practice evaluating definite integrals involving logarithmic functions
  • Explore advanced techniques in calculus, such as substitution methods
  • Learn about the properties of logarithms and their applications in calculus
USEFUL FOR

Students studying calculus, educators teaching integration techniques, and anyone looking to strengthen their understanding of definite integrals and logarithmic functions.

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Here is the question:

Calculus integral help?

Evaluate: the integral from 2(top) to 1(bottom) and the function is: x^2(lnx) dx

Here is a link to the question:

Calculus integral help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Henry,

We are given to evaluate:

$\displaystyle \int_1^2x^2\ln(x)\,dx$

Using integration by parts, we may let:

$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$

$\displaystyle dv=x^2\,dx\,\therefore\,v=\frac{1}{3}x^3$

and so we have:

$\displaystyle \int_1^2x^2\ln(x)\,dx=\left[\frac{1}{3}x^3\ln(x) \right]_1^2-\frac{1}{3}\int_1^2x^2\,dx$

$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)-\frac{1}{3}\left[\frac{1}{3}x^3 \right]_1^2$

$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)-\frac{1}{9}\left(8-1 \right)$

$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)-\frac{7}{9}$
 

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