Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)

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First, rewrite the integral to clearly identify the integrand and the limits: \(\int_{-1}^{0} \frac{6}{\sqrt{3 - 2t - t^{2}}} \, dt\).

Complete the square inside the square root in the denominator to simplify the expression. Start with the quadratic expression: \(3 - 2t - t^{2}\). Rewrite it as \(-(t^{2} + 2t - 3)\), then complete the square for \(t^{2} + 2t - 3\).

Express \(t^{2} + 2t - 3\) in the form \((t + a)^{2} + b\) by finding \(a\) and \(b\). Recall that \(t^{2} + 2t = (t + 1)^{2} - 1\), so \(t^{2} + 2t - 3 = (t + 1)^{2} - 4\).

Substitute back into the integral, so the denominator becomes \(\sqrt{-( (t + 1)^{2} - 4 )} = \sqrt{4 - (t + 1)^{2}}\). This suggests a trigonometric substitution of the form \(t + 1 = 2 \sin \theta\).

Change the variable of integration using the substitution \(t + 1 = 2 \sin \theta\), find \(dt\) in terms of \(d\theta\), adjust the limits accordingly, and rewrite the integral in terms of \(\theta\). Then, simplify and integrate using the standard integral formula for \(\int \frac{d\theta}{\sqrt{1 - \sin^{2} \theta}}\) or equivalent.

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주요 개념

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Definite Integral

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫ from a to b of f(t) dt, where a and b are the lower and upper bounds. Evaluating a definite integral results in a numerical value representing the accumulated quantity over that interval.

Integration of Rational Functions Involving Square Roots

Integrals involving expressions like 1/√(quadratic) often require algebraic manipulation such as completing the square. This transforms the integrand into a form suitable for standard integral formulas involving inverse trigonometric functions or logarithms, facilitating easier evaluation.

Completing the Square

Completing the square rewrites a quadratic expression ax² + bx + c into the form (x + d)² + e. This technique simplifies the integrand, especially under square roots, enabling the use of standard integral formulas. It is essential for transforming the integral into a recognizable and solvable form.

Evaluate the integrals in Exercises 77–90.79. ∫(from -1 to 0)6dt/√(3-2t-t²)