Evaluating integrals Evaluate the following integrals.

∫₁⁴ ((√v + v) / v ) dv

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Step 1: Simplify the integrand. The given integral is ∫₁⁴ ((√v + v) / v) dv. Break the fraction into two terms: (√v / v) + (v / v). Simplify each term to get (1/√v) + 1.

Step 2: Rewrite the integrand in terms of exponents. Recall that √v = v^(1/2), so 1/√v = v^(-1/2). The integrand becomes v^(-1/2) + 1.

Step 3: Apply the power rule for integration. The integral of v^n is (v^(n+1))/(n+1), provided n ≠ -1. Integrate each term separately: ∫v^(-1/2) dv and ∫1 dv.

Step 4: Compute the antiderivatives. For v^(-1/2), the antiderivative is (v^(1/2))/(1/2) = 2√v. For 1, the antiderivative is v. Combine these results to get the antiderivative: 2√v + v.

Step 5: Evaluate the definite integral. Substitute the limits of integration (1 and 4) into the antiderivative, 2√v + v, and compute the difference between the values at the upper and lower limits.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration

Integration is a fundamental concept in calculus that involves finding the accumulated area under a curve represented by a function. It is the reverse process of differentiation and can be used to calculate quantities such as areas, volumes, and total accumulated change. The integral symbol (∫) denotes the operation, and definite integrals have specified limits, indicating the interval over which the integration is performed.

Simplifying Expressions

Simplifying expressions is a crucial step in calculus that involves rewriting mathematical expressions in a more manageable form. In the context of integrals, this often includes factoring, combining like terms, or reducing fractions to make the integration process easier. For the given integral, simplifying the expression ((√v + v) / v) can help in breaking it down into simpler components that are easier to integrate.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration, stating that if a function is continuous on an interval, then the integral of its derivative over that interval equals the change in the function's values at the endpoints. This theorem provides a method for evaluating definite integrals by finding an antiderivative of the integrand, which is essential for solving the integral in the question.

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Related Practice

Textbook Question

Average value of the derivative Suppose ƒ ' is a continuous function for all real numbers. Show that the average value of the derivative on an interval [a, b] is ƒ⁻' = (ƒ(b) ―ƒ(a))/ (b―a) . Interpret this result in terms of secant lines.

Textbook Question

Variations on the substitution method Evaluate the following integrals.

∫ y²/(y + 1)⁴ dy

Textbook Question

Consider the function

ƒ(t) = { t if ―2 ≤ t < 0

t²/2 if 0 ≤ t ≤ 2

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(f) Find a constant C such that F(𝓍) = G(𝓍) + C .

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Textbook Question

Evaluating integrals Evaluate the following integrals.