Evaluating integrals Evaluate the following integrals.
β«βΒ² (2π + 1)Β³ dπ
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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.
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.
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.