Let f(x)=1/(x-2)^2. Which setup for ∫0^6 f(x) dx uses the required independent one-sided limits?

• lim(t -> 6 from below) ∫0^t f(x) dx, keeping the interior singularity inside each interval
• lim(u -> 2 from below) ∫0^u f(x) dx + lim(v -> 2 from above) ∫v^6 f(x) dx
• lim(a -> 0 from above) ∫a^6 f(x) dx, approaching the regular left endpoint instead
• lim(h -> 0 from above) [∫0^(2-h) f(x) dx + ∫(2+h)^6 f(x) dx], requiring the cutoffs to move together

Let f be continuous on [1, infinity). Which limit must be tested for its improper integral on this interval?

• Substitute infinity into f(x) and call that value the integral
• Take lim(t -> infinity) of the integral from 1 to t
• Take lim(t -> 1 from above) of the integral from 1 to t

For which p does the integral from 1 to infinity of 1/x^p dx converge?

• p < 1
• p = 1
• p > 1

Evaluate the integral from 1 to infinity of 1/x^2 dx.

Evaluate the integral from 0 to 1 of 1/sqrt(x) dx.

Let f(x)=x/(1+x^2). For every t>0, ∫-t^t f(x) dx=0, while ∫0^b f(x) dx=ln(1+b^2)/2. What follows about ∫-infinity^infinity f(x) dx?

• It diverges because the right-hand limit is not finite.
• It converges to 0 because every symmetric cutoff gives 0.
• It converges to 1 because each denominator contains 1.
• It has value 0 because opposite infinite tails always cancel.

For real p>1, ∫1^infinity 1/x^p dx has value 3. What is p?

• p = 1/3
• p = 3/2
• p = 4/3
• p = 5/3

Evaluate the integral from 1 to infinity of 1/x^3 dx. Enter a decimal.

A friend says ∫1^infinity 1/x dx is finite because 1/x tends to 0. Which check settles this claim?

• Since 1/x never equals 0, every finite interval has area 0
• The integral to t is ln(t), which grows without bound
• Since 1/x tends to 0, its integral must have the same limit

Let f be continuous on the real line. If ∫-infinity^0 f(x) dx and ∫0^infinity f(x) dx each converge, then ∫-infinity^infinity f(x) dx converges.