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91Ó°ÊÓ

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

∫23lnxxdx

Short Answer

Expert verified

∫23lnxxdx=12[ln23-ln22].

Step by step solution

01

Step 1. Given information. 

A definite integral is given as∫23lnxxdx.

02

Step 2. Using the Fundamental theorem of Calculus.

Let

f(x)=x2g(x)=lnx

then

f'(x)=2xg'(x)=1xf'(g(x))=2(lnx)

Now we get

∫23lnxxdx=12∫232lnxxdx=12[ln2x]23[f'(g(x))g'(x)dx=f(g(x))]=12[ln23-ln22]

The exact value of the given definite integral is 12[ln23-ln22].

03

Step 3. The graph to verify the answer is

The solution is area under graph which is

a=0.36324797≈12[ln23-ln22]

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