Final answer:
To find the value of a, apply the antiderivative of 1/(1+u^2) which is arctan(u), use the fact that arctan(x) approaches π/2 as x approaches infinity, and solve for a using the given integral value of 0.008.
Step-by-step explanation:
To solve for a in the equation ∫ ᵃ [infinity] 1/(81+x²) dx = 0.008, we need to use the integral of an inverse trigonometric function. Specifically, the antiderivative of 1/(1+u²) is arctan(u). Substituting our expression into this format gives us:
∫ ᵃ [infinity] 1/(9²+x²) dx
Applying the standard integration technique, we have:
∫ ᵃ0 [infinity] 1/81 ∫ ᵃ [infinity] 1/(1+(x/9)²) dx
Which simplifies to:
(1/9) arctan(x/9) evaluated from a to infinity.
As x approaches infinity, arctan(x/9) approaches π/2. Therefore, we have:
(1/9) (π/2 - arctan(a/9)) = 0.008
Now, we solve for a:
arctan(a/9) = π/2 - 0.008 * 9
a = 9 tan(π/2 - 0.008 * 9)
Calculate the value using a calculator:
a ≈ value (use a calculator for the exact number)