Textbook Question
7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
26. β«[β2 to β2] β(xΒ² - 1)/x dx
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8. Definite Integrals
Substitution
2:51 minutesProblem 5.R.57
Textbook Question
Evaluating integrals Evaluate the following integrals.
β«βΒ² (2π + 1)Β³ dπ
Verified step by step guidance1
Step 1: Recognize that the integral β«βΒ² (2π + 1)Β³ dπ involves a polynomial raised to a power. To simplify, consider using substitution. Let u = 2π + 1, which transforms the integral into a simpler form.
Step 2: Compute the derivative of u with respect to π: du/dπ = 2. Rearrange to express du in terms of dπ: du = 2 dπ.
Step 3: Adjust the limits of integration to match the substitution. When π = 0, u = 2(0) + 1 = 1. When π = 2, u = 2(2) + 1 = 5. The integral now becomes β«ββ΅ uΒ³ (1/2) du, where the factor 1/2 comes from the substitution for dπ.
Step 4: Factor out the constant 1/2 from the integral: (1/2) β«ββ΅ uΒ³ du. Now, apply the power rule for integration: β« uβΏ du = (uβΏβΊΒΉ)/(n+1) + C, where n is the exponent.
Step 5: Use the power rule to integrate uΒ³: (1/2) [(uβ΄)/4] evaluated from u = 1 to u = 5. Substitute the limits into the result to complete the evaluation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral represents the signed area under a curve defined by a function over a specific interval. In this case, the integral β«βΒ² (2π + 1)Β³ dπ calculates the area between the curve (2π + 1)Β³ and the x-axis from x = 0 to x = 2. The limits of integration (0 and 2) indicate the bounds of the area being evaluated.
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Integration Techniques
To evaluate integrals, various techniques can be employed, such as substitution, integration by parts, or polynomial expansion. For the integral β«βΒ² (2π + 1)Β³ dπ, expanding the integrand (2π + 1)Β³ into a polynomial form can simplify the integration process, making it easier to compute the definite integral.
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Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then β«βα΅ f(x) dx = F(b) - F(a). This theorem allows us to evaluate definite integrals by finding an antiderivative of the integrand, which is essential for solving the given integral problem.
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