Area of the region bounded between two curves with integration by parts

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SUMMARY

The discussion revolves around calculating the area bounded between the curves y=34ln(x) and y=xln(x) using integration by parts. The initial attempt incorrectly set the lower bound of integration at 0, which is invalid as neither function is defined at that point. The correct bounds for the integral are from 1 to 34. The final integral evaluated is ∫(34ln(x) - xln(x))dx, leading to the area calculation of 1156ln(34) - 1156 - (1156ln(34)/2) + 289.

PREREQUISITES
  • Understanding of integration by parts, specifically the formula ∫udv = uv - ∫vdu.
  • Familiarity with logarithmic functions, particularly natural logarithms.
  • Ability to find intersection points of curves to determine integration limits.
  • Basic knowledge of evaluating definite integrals.
NEXT STEPS
  • Review the concept of integration by parts in calculus.
  • Study the properties and behavior of logarithmic functions.
  • Learn how to find intersection points of curves analytically.
  • Practice evaluating definite integrals with various bounds.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques and area calculations between curves. This discussion is also beneficial for educators seeking to clarify common mistakes in integration problems.

maladroit
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Homework Statement



Find the area bounded between the two curves
y=34ln(x) and y=xln(x)


Homework Equations



Integration by parts: \intudv= uv-\intvdu

The Attempt at a Solution


First I found the intersection points of the two equation to set the upper and lower bounds. The lower bound is 0 and the upper bound is 34. My integrand is as follows--

\int34ln(x)-xln(x) with the limits of integration being from 0 to 34.

I evaluated the integral using integration by parts, and eventually came up with the following solution...

\int34ln(x)-xln(x)= 34xln(x)-34x-(x^{2}ln(x)/2) -(1/4)x^{2}

and evaluated from 0 to 34, the answer is 1156ln(34)-1156-(1156ln(34)/2)+289

I am not quite sure where my mistake is being made. I verified my answer using a graphing calculator (although that does not absolutely make my answer correct), so if anyone sees where I am making my mistake I would greatly appreciate the help!
 
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When you say the lower bound is 0, I don't see how that is possible. From your integral, it appears that you are integrating with respect to x. Neither of your functions is defined at x = 0.
 
check this again

34xln(x)-34x-(xLaTeX Code: ^{2} ln(x)/2) -(1/4)xLaTeX Code: ^{2}

one of the sign is wrong
 
My mistake was in the bounds, which should have been from 1 to 34. Thank you for your help!
 

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