Evaluating integrals Evaluate the following integrals.
β«βΒΉ βπ (βπ + 1) dπ
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Step 1: Begin by expanding the integrand. The given integral is β«βΒΉ βπ (βπ + 1) dπ. Distribute βπ across the terms inside the parentheses to rewrite the integrand as β«βΒΉ (π + βπ) dπ.
Step 2: Split the integral into two separate integrals for easier computation: β«βΒΉ (π + βπ) dπ = β«βΒΉ π dπ + β«βΒΉ βπ dπ.
Step 3: Evaluate the first integral, β«βΒΉ π dπ. Use the power rule for integration, which states β«πβΏ dπ = (πβΏβΊΒΉ)/(n+1) + C, where n β -1. Here, n = 1, so β«βΒΉ π dπ = [πΒ²/2]βΒΉ.
Step 4: Evaluate the second integral, β«βΒΉ βπ dπ. Rewrite βπ as π^(1/2) and apply the power rule for integration. For n = 1/2, β«π^(1/2) dπ = (π^(3/2))/(3/2) + C. Simplify to (2/3)π^(3/2). Evaluate this expression from 0 to 1: [(2/3)π^(3/2)]βΒΉ.
Step 5: Combine the results of the two integrals. Add the evaluated results of β«βΒΉ π dπ and β«βΒΉ βπ dπ to obtain the final value of the integral β«βΒΉ βπ (βπ + 1) dπ.
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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 between two specified limits. In this case, the integral β«βΒΉ βπ (βπ + 1) dπ is evaluated from 0 to 1, which means we are calculating the area under the curve of the function βπ (βπ + 1) from x = 0 to x = 1.
To evaluate integrals, various techniques can be employed, such as substitution, integration by parts, or recognizing patterns. For the given integral, simplifying the integrand βπ (βπ + 1) may involve expanding the expression or using substitution to make the integration process more manageable.
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 is essential for evaluating definite integrals, as it allows us to find the area under the curve by calculating the difference of the antiderivative at the upper and lower limits.