7–64. Integration review Evaluate the following integrals.
18. ∫ from 3 to 7 of (t - 6) * √(t - 3) dt

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Recognize that the integral is a definite integral from 3 to 7 of the function \((t - 6) \cdot \sqrt{t - 3}\, dt\). The goal is to evaluate this integral step by step.

Simplify the integrand \((t - 6) \cdot \sqrt{t - 3}\). Let \(u = t - 3\), which implies \(t = u + 3\) and \(dt = du\). Also, note that when \(t = 3\), \(u = 0\), and when \(t = 7\), \(u = 4\).

Substitute \(t = u + 3\) into the integrand. The expression \((t - 6)\) becomes \((u + 3 - 6) = (u - 3)\), and \(\sqrt{t - 3}\) becomes \(\sqrt{u}\). The integral now becomes \(\int_{0}^{4} (u - 3) \cdot \sqrt{u} \, du\).

Distribute \(\sqrt{u}\) across \(u - 3\) to rewrite the integrand as \(u^{3/2} - 3u^{1/2}\). The integral becomes \(\int_{0}^{4} u^{3/2} \, du - 3 \int_{0}^{4} u^{1/2} \, du\).

Evaluate each term separately. Use the power rule for integration: \(\int u^n \, du = \frac{u^{n+1}}{n+1} + C\). For \(\int u^{3/2} \, du\), \(n = 3/2\), and for \(\int u^{1/2} \, du\), \(n = 1/2\). After finding the antiderivatives, apply the limits of integration (from 0 to 4) to compute the definite integral.

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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. It is denoted as ∫ from a to b of f(x) dx, where 'a' and 'b' are the lower and upper limits, respectively. The result of a definite integral is a numerical value that quantifies the accumulation of the function's values over the interval [a, b].

Integration Techniques

Integration techniques are methods used to evaluate integrals, especially when they cannot be solved using basic antiderivatives. Common techniques include substitution, integration by parts, and trigonometric substitution. In the given integral, substitution may simplify the expression, making it easier to evaluate the integral.

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 ∫ from a to b of f(x) dx = F(b) - F(a). This theorem allows us to evaluate definite integrals by finding the antiderivative and calculating its values at the limits of integration.

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