Question

Evaluate the improper integrals in Exercises 53–62.∫ from −∞ to ∞ of (1 / (4x² + 9)) dx

Accepted Answer

Recognize that the integral is an improper integral because the limits of integration are from $$-\infty to$$ $$\infty. $$This means we need to evaluate the integral as a limit: $$\int_{-\infty}^{\infty} \frac{1}{4x$$^{2}$$ + 9} \, dx = \lim_{A 	o \infty} \int_{-A}^{A} \frac{1}{4x$$^{2}$$ + 9} \, dx. $$Identify the integrand as a rational function that resembles the form $$\frac{1}{x$$^{2}$$ + a$$^{2}$$}$$, which suggests using a standard integral formula: $$\int \frac{dx}{x$$^{2}$$ + a$$^{2}$$} = \frac{1}{a} \arctan\left(\frac{x}{a}ight) + C. $$Rewrite the denominator to match the standard form by factoring out the constant: $$4x^{2} + 9$$ = $$(2x)^{2}$$ + $$3^{2}. $$Let $$u = 2x$$, so $$du = 2 dx or$$ $$dx = \frac{du}{2}. $$Substitute into the integral: $$\int \frac{1}{u$$^{2}$$ + 3$$^{2}$$} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{u$$^{2}$$ + 3$$^{2}$$}. $$Then apply the arctangent formula to get $$\frac{1}{2} \cdot \frac{1}{3} \arctan\left(\frac{u}{3}ight) + C = \frac{1}{6} \arctan\left(\frac{2x}{3}ight) + C. $$Evaluate the definite integral from $$-A to$$ $$A$$ and then take the limit as $$A 	o \infty$$: $$\lim_{A 	o \infty} \left[ \frac{1}{6} \arctan\left(\frac{2A}{3}ight) - \frac{1}{6} \arctan\left(\frac{-2A}{3}ight) ight]. $$Use the fact that $$\arctan(x)$$ approaches $$\frac{\pi}{2} as$$ $$x 	o \infty$$ and $$-\frac{\pi}{2} as$$ $$x 	o -\infty to $$simplify.

Evaluate the improper integrals in Exercises 53–62.
∫ from −∞ to ∞ of (1 / (4x² + 9)) dx

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Improper Integrals

Integration of Rational Functions

Use of Limits in Definite Integrals