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Magazine Chapter 25: Problem 31
evaluate the given definite integrals. $$\int_{0}^{1}\left(x^{2}+3\right)\left(x^{3}+9 x+6\right)^{2} d x$$
Short Answer
Expert verified
The integral evaluates to \( \frac{4078}{9} \).
Step by step solution
01
Set up the integrand
Identify the function to be integrated: \[ \int_{0}^{1}\left(x^{2}+3\right)\left(x^{3}+9x+6\right)^{2} \, dx \] This integral involves a product of functions.
02
Expand the inner function
First, expand \( (x^3 + 9x + 6)^2 \): This requires multiplying it out: \[(x^3 + 9x + 6)(x^3 + 9x + 6) = x^6 + 18x^4 + 36x^3 + 81x^2 + 108x + 36 \]
03
Expand the integrand
Multiply \( x^2 + 3 \) by the expanded function to simplify \[(x^2 + 3)(x^6 + 18x^4 + 36x^3 + 81x^2 + 108x + 36) \]Multiply each term separately:- \(x^2 \cdot x^6 = x^8\)- \(x^2 \cdot 18x^4 = 18x^6\)- \(x^2 \cdot 36x^3 = 36x^5\)- \(x^2 \cdot 81x^2 = 81x^4\)- \(x^2 \cdot 108x = 108x^3\)- \(x^2 \cdot 36 = 36x^2\)- \(3 \cdot x^6 = 3x^6\)- \(3 \cdot 18x^4 = 54x^4\)- \(3 \cdot 36x^3 = 108x^3\)- \(3 \cdot 81x^2 = 243x^2\)- \(3 \cdot 108x = 324x\)- \(3 \cdot 36 = 108\)Summing up all terms gives:\[ x^8 + 21x^6 + 36x^5 + 135x^4 + 216x^3 + 279x^2 + 324x + 108 \]
04
Integrate term by term
Now, integrate each term with respect to \(x\) from 0 to 1:- \( \int x^8 \, dx = \frac{x^9}{9} \)- \( \int 21x^6 \, dx = \frac{21x^7}{7} = 3x^7 \)- \( \int 36x^5 \, dx = \frac{36x^6}{6} = 6x^6 \)- \( \int 135x^4 \, dx = \frac{135x^5}{5} = 27x^5 \)- \( \int 216x^3 \, dx = \frac{216x^4}{4} = 54x^4 \)- \( \int 279x^2 \, dx = \frac{279x^3}{3} = 93x^3 \)- \( \int 324x \, dx = \frac{324x^2}{2} = 162x^2 \)- \( \int 108 \, dx = 108x \)This results in the antiderivative:\[ \frac{x^9}{9} + 3x^7 + 6x^6 + 27x^5 + 54x^4 + 93x^3 + 162x^2 + 108x \]
05
Evaluate the antiderivative
Plug in the upper limit (1) and the lower limit (0) into the integrated function:At \(x = 1\): \[ \frac{1^9}{9} + 3(1^7) + 6(1^6) + 27(1^5) + 54(1^4) + 93(1^3) + 162(1^2) + 108(1) = \frac{1}{9} + 3 + 6 + 27 + 54 + 93 + 162 + 108 \]This simplifies to a total sum of:\[ \frac{1}{9} + 453 \]At \(x = 0\), the antiderivative evaluates to 0.Therefore, the definite integral equals \[ \frac{1}{9} + 453 = \frac{1}{9} + \frac{4077}{9} = \frac{4078}{9} \]
06
Conclusion: Write the final answer
The evaluation of the definite integral is \( \frac{4078}{9} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Step-by-Step Integration
Calculus can be intimidating, but breaking down integration into step-by-step processes makes things manageable. To evaluate definite integrals like the one in our exercise, we start by analyzing the given function carefully. Here, we have a product of two functions:
- \( x^2 + 3 \)
- \( (x^3 + 9x + 6)^2 \)
Polynomial Expansion
When dealing with expressions such as \( (x^3 + 9x + 6)^2 \), expansion is a necessary step. Polynomial expansion involves multiplying terms in the expression to reach a simplified polynomial. This allows us to handle each term separately during integration.We multiply each term of the first polynomial by every term of the second:
- Multiply: \( x^3 \times x^3 = x^6 \)
- Continue for all combinations.
Antiderivative Evaluation
The antiderivative, or indefinite integral, is a crucial part of integral calculus. Once we have our simplified polynomial, finding the antiderivative means you have to integrate term by term. This process involves applying basic integration rules:
- For \( x^a \), the antiderivative is \( \frac{x^{a+1}}{a+1} \)
- For constants, multiply by \( x \)
Integral Calculus
Integral calculus enables us to determine areas under curves and solve accumulation problems. In definite integrals, the focus is finding the net area between the curve and the x-axis between two points.
After finding the antiderivative, evaluate it at the boundaries of the given interval. In our case, substituting 1 and 0 into the antiderivative gives us the net change from the starting to ending point.
- Plug in upper bounds.
- Subtract the result from plugging in lower bounds.
