$\int 6 x e^{2 x} d x = \square +C
Answer (2)
To solve ∫ 6 x e 2 x d x , we utilize integration by parts, choosing u = 6 x and d v = e 2 x d x . After calculating d u and v , we apply the integration by parts formula to arrive at the answer: 2 1 e 2 x ( 6 x − 3 ) + C .
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Recognize the need for integration by parts due to the product of two functions.
Choose u = 6 x and d v = e 2 x d x , then find d u = 6 d x and v = 2 1 e 2 x .
Apply the integration by parts formula: ∫ u d v = uv − ∫ v d u , resulting in ∫ 6 x e 2 x d x = 3 x e 2 x − 3 ∫ e 2 x d x .
Evaluate the remaining integral and simplify to get the final answer: 2 1 e 2 x ( 6 x − 3 ) .
Explanation
Recognizing Integration by Parts We are asked to find the indefinite integral of 6 x e 2 x . This calls for integration by parts, a technique useful when integrating the product of two functions. The formula for integration by parts is: ∫ u d v = uv − ∫ v d u
Choosing u and dv We need to choose which part of our integrand will be 'u' and which will be 'dv'. A good strategy is to choose 'u' such that its derivative simplifies the integral. Let's choose: u = 6 x and d v = e 2 x d x
Finding du and v Now we find 'du' and 'v': d u = d x d ( 6 x ) d x = 6 d x v = ∫ e 2 x d x = 2 1 e 2 x
Applying Integration by Parts Now we plug these into the integration by parts formula: ∫ 6 x e 2 x d x = ( 6 x ) ( 2 1 e 2 x ) − ∫ ( 2 1 e 2 x ) ( 6 d x ) ∫ 6 x e 2 x d x = 3 x e 2 x − 3 ∫ e 2 x d x
Evaluating the Remaining Integral Now we need to evaluate the remaining integral: ∫ e 2 x d x = 2 1 e 2 x + C 1
Final Substitution and Simplification Substitute this back into our expression: ∫ 6 x e 2 x d x = 3 x e 2 x − 3 ( 2 1 e 2 x ) + C ∫ 6 x e 2 x d x = 3 x e 2 x − 2 3 e 2 x + C
Simplifying the Expression We can factor out 2 1 e 2 x to simplify: ∫ 6 x e 2 x d x = 2 1 e 2 x ( 6 x − 3 ) + C
Final Answer Thus, the indefinite integral of 6 x e 2 x is: 2 1 e 2 x ( 6 x − 3 )
Examples
Imagine you're analyzing the growth of a population where the growth rate is proportional to both the current population size and time. The integral you solved is similar to those used in such models, where you might need to find the total population growth over a period. By understanding how to integrate functions like 6 x e 2 x , you can model and predict various real-world phenomena involving exponential growth and decay, such as compound interest, radioactive decay, or the spread of a disease.
