Question

Evaluate the improper integrals in Exercises 53–62.∫ from 0 to 3 of (1 / √(9 − x²)) dx

Accepted Answer

Recognize that the integral $$ \int_0^3 \frac{1}{\sqrt{9 - x^2}} \, dx  is an $$improper integral because the integrand involves a square root in the denominator that becomes zero at the upper limit $$ x = 3 . $$This means the function approaches infinity there, so we need to treat the integral as a limit. Rewrite the integral as a limit: $$ \lim_{t 	o 3^-} \int_0^t \frac{1}{\sqrt{9 - x^2}} \, dx . $$This allows us to evaluate the integral over $$ [0, t] $$ where $$ t \u003c 3 $$ and then take the limit as $$ t $$ approaches 3 from the left. Recall the antiderivative formula for $$ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left( \frac{x}{a} ight) + C . In $$this problem, $$ a = 3 $$, so the antiderivative is $$ \arcsin\left( \frac{x}{3} ight) + C . $$Evaluate the definite integral from 0 to $$ t $$ using the antiderivative: $$ \int_0^t \frac{1}{\sqrt{9 - x^2}} \, dx = \arcsin\left( \frac{t}{3} ight) - \arcsin(0) . $$Since $$ \arcsin(0) = 0 $$, this simplifies to $$ \arcsin\left( \frac{t}{3} ight) . $$Finally, take the limit as $$ t 	o 3^- $$: $$ \lim_{t 	o 3^-} \arcsin\left( \frac{t}{3} ight) . $$This will give the value of the improper integral.

Evaluate the improper integrals in Exercises 53–62.
∫ from 0 to 3 of (1 / √(9 − x²)) dx

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Key Concepts

Improper Integrals

Integration of Functions Involving Square Roots

Trigonometric Substitution