Apply integration by parts with u = ln ( 3 x ) and d v = x 2 d x , leading to d u = x 1 d x and v = 3 x 3 .
Use the integration by parts formula: ∫ u d v = uv − ∫ v d u , resulting in ∫ 1 3 x 2 ln 3 x d x = [ 3 x 3 ln 3 x ] 1 3 − ∫ 1 3 3 x 2 d x .
Evaluate the integral and the term separately: ∫ 1 3 3 x 2 d x = 9 26 and [ 3 x 3 ln 3 x ] 1 3 = 3 53 ln 3 .
Combine the results to get the final answer: 3 53 ln 3 − 9 26 .
Explanation
Problem Setup We are asked to find the exact value of the definite integral ∫ 1 3 x 2 ln ( 3 x ) d x and express it in the form a ln b + c , where a and c are rational numbers and b is an integer.
Integration by Parts To solve this integral, we will use integration by parts. The formula for integration by parts is ∫ u d v = uv − ∫ v d u . We choose u = ln ( 3 x ) and d v = x 2 d x .
Finding du and v Now we find d u and v . If u = ln ( 3 x ) , then d u = 3 x 1 "."3 d x = x 1 d x . If d v = x 2 d x , then v = ∫ x 2 d x = 3 x 3 .
Applying the Formula Applying the integration by parts formula, we have: ∫ 1 3 x 2 ln ( 3 x ) d x = [ 3 x 3 ln ( 3 x ) ] 1 3 − ∫ 1 3 3 x 3 ⋅ x 1 d x = [ 3 x 3 ln ( 3 x ) ] 1 3 − ∫ 1 3 3 x 2 d x
Evaluating the Integral Now we evaluate the integral ∫ 1 3 3 x 2 d x :
∫ 1 3 3 x 2 d x = 3 1 ∫ 1 3 x 2 d x = 3 1 [ 3 x 3 ] 1 3 = 3 1 ( 3 3 3 − 3 1 3 ) = 3 1 ( 3 27 − 3 1 ) = 3 1 ⋅ 3 26 = 9 26
Evaluating the Term Next, we evaluate the term [ 3 x 3 ln ( 3 x ) ] 1 3 :
[ 3 x 3 ln ( 3 x ) ] 1 3 = 3 3 3 ln ( 3 ⋅ 3 ) − 3 1 3 ln ( 3 ⋅ 1 ) = 3 27 ln ( 9 ) − 3 1 ln ( 3 ) = 9 ln ( 9 ) − 3 1 ln ( 3 ) = 9 ln ( 3 2 ) − 3 1 ln ( 3 ) = 18 ln ( 3 ) − 3 1 ln ( 3 ) = 3 54 ln ( 3 ) − 3 1 ln ( 3 ) = 3 53 ln ( 3 )
Combining the Results Combining the results, we have: ∫ 1 3 x 2 ln ( 3 x ) d x = 3 53 ln ( 3 ) − 9 26
Final Answer The exact value of the definite integral is 3 53 ln 3 − 9 26 . This is in the form a ln b + c , where a = 3 53 , b = 3 , and c = − 9 26 .
Examples
Imagine you're calculating the total energy produced by a solar panel over a period of time, where the energy production rate is described by a function involving a logarithmic term. Integrals like this help determine the accumulated energy output, crucial for assessing the panel's efficiency and overall energy yield. Understanding integration by parts allows for accurate modeling and prediction of energy-related phenomena.