Improper Integrals Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
Evaluate โˆซ011xโ€‰dx\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx (Type II).

Solution

  1. 1
    The integrand xโˆ’1/2x^{-1/2} is unbounded at x=0x = 0, so replace the lower limit with a>0a > 0 and take the limit.
  2. 2
    Integrate xโˆ’1/2x^{-1/2}: the antiderivative is 2x2\sqrt{x}. Evaluate the definite integral: limโกaโ†’0+[2x]a1=limโกaโ†’0+(21โˆ’2a)\lim_{a\to0^+}\left[2\sqrt{x}\right]_a^1 = \lim_{a\to0^+}(2\sqrt{1} - 2\sqrt{a})
  3. 3
    Take the limit: 2aโ†’02\sqrt{a} \to 0 as aโ†’0+a \to 0^+, so the integral converges to 2โˆ’0=22 - 0 = 2.

Answer

22
Replace the problematic endpoint with aโ†’0+a \to 0^+. Since aโ†’0\sqrt{a} \to 0, the result is finite.

About Improper Integrals

Integrals where the interval of integration is infinite (Type I: โˆซaโˆžf(x)โ€‰dx\int_a^{\infty} f(x)\,dx) or the integrand has an infinite discontinuity on the interval (Type II: โˆซabf(x)โ€‰dx\int_a^b f(x)\,dx where ff blows up at some point in [a,b][a, b]). Evaluated as limits of proper integrals.

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