Improper Integrals (Edexcel A-Level Further Mathematics): Revision Notes

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5.2.1 Improper Integrals

What Are Improper Integrals?

An improper integra****l arises when:

  1. The integrand is undefined at one or more points within the range of integration.
  2. The range of integration extends to infinity. Despite these issues, improper integrals can often be evaluated by taking limits. For instance, integrals involving functions with vertical asymptotes or those defined over infinite intervals are common examples.

How to Evaluate Improper Integrals

Undefined Integrand at a Point (Type I):

For example, consider:

∫021x dx\int_0^2 \frac{1}{\sqrt{x}} \, dx

The integrand 1x\frac{1}{\sqrt{x}} becomes undefined at x = 0.

Solution Method:

Replace the problematic limit with a parameter tt, then evaluate the limit:

∫021x dx=lim⁡t→0+∫t21x dx\int_0^2 \frac{1}{\sqrt{x}} \, dx = \lim_{t \to 0^+} \int_t^2 \frac{1}{\sqrt{x}} \, dx

Infinite Limits of Integration (Type II):

For example, consider:

∫1∞1x2 dx\int_1^\infty \frac{1}{x^2} \, dx

Here, the upper limit extends to infinity.

Solution Method:

Replace ∞\infty with a parameter tt, then take the limit:

∫1∞1x2 dx=lim⁡t→∞∫1t1x2 dx\int_1^\infty \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2} \, dx

Worked Examples

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Example 1: Evaluate ∫1∞1x2 dx\int_1^\infty \frac{1}{x^2} \, dx

Step 1**: Set up the limit:**

Replace the infinite limit with a parameter tt:

∫1∞1x2 dx=lim⁡t→∞∫1t1x2 dx\int_1^\infty \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2} \, dx

Step 2**: Integrate:**

The antiderivative of 1x2\frac{1}{x^2} is 1x\frac{1}{x}:

∫1t1x2 dx=[−1x]1t=−1t+11\int_1^t \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^t = -\frac{1}{t} + \frac{1}{1}

Step 3: Take the limit:

As t→∞t \to \infty,1t→0 \frac{1}{t} \to 0:

lim⁡t→∞(−1t+1)=0+1=1\lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 0 + 1 = 1

Result:

∫1∞1x2 dx=:success[1]\int_1^\infty \frac{1}{x^2} \, dx = :success[1]
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Example 2: Evaluate ∫021x dx\int_0^2 \frac{1}{\sqrt{x}} \, dx

Step 1: Set up the limit:

The integrand is undefined at x = 0.

Replace this limit with tt:

∫021x dx=lim⁡t→0+∫t21x dx\int_0^2 \frac{1}{\sqrt{x}} \, dx = \lim_{t \to 0^+} \int_t^2 \frac{1}{\sqrt{x}} \, dx

Step 2: Integrate:

The antiderivative of 1x\frac{1}{\sqrt{x}} is 2x2\sqrt{x}:

∫t21x dx=[2x]t2=22−2t\int_t^2 \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_t^2 = 2\sqrt{2} - 2\sqrt{t}

Step 3: Take the limit:

As t→0+t \to 0^+, t→0\sqrt{t} \to 0

lim⁡t→0+(22−2t)=22\lim_{t \to 0^+} \left( 2\sqrt{2} - 2\sqrt{t} \right) = 2\sqrt{2}

Result:

∫021x dx=:success[22]\int_0^2 \frac{1}{\sqrt{x}} \, dx = :success[2\sqrt{2}]
lightbulbExample

Example 3: Evaluate ∫0∞e−x dx\int_0^\infty e^{-x} \, dx

Step 1: Set up the limit:

The upper limit is infinite:

∫0∞e−x dx=lim⁡t→∞∫0te−x dx\int_0^\infty e^{-x} \, dx = \lim_{t \to \infty} \int_0^t e^{-x} \, dx

Step 2: Integrate:

The antiderivative of e−xe^{-x} is −e−x−e^{-x}:

∫0te−x dx=[−e−x]0t=−(e−t−e0)\int_0^t e^{-x} \, dx = \left[ -e^{-x} \right]_0^t = -(e^{-t} - e^0)

Step 3: Take the limit:

As t→∞t \to \infty, e−t→0e^{-t} \to 0:

lim⁡t→∞(−e−t+1)=0+1=1\lim_{t \to \infty} \left( -e^{-t} + 1 \right) = 0 + 1 = 1

Result:

∫0∞e−x dx=:success[1]\int_0^\infty e^{-x} \, dx = :success[1]
lightbulbExample

Example 4:

∫1∞1x2 dx\int_{1}^{\infty} \frac{1}{x^2} \, dx

Step 1: Replace the undefined limit with a constant, say t, and perform the integration:

∫1tx−2 dx=[−x−1]1t=−(t−1+1)\int_{1}^{t} x^{-2} \, dx = \left[ -x^{-1} \right]_{1}^{t} = -(t^{-1} + 1)

Step 2: Show the process of tt approaching the undefined limit:

lim⁡t→∞(−t−1+1)=lim⁡t→∞(−t−1)⏟=0+1=0+1=:success[1]\lim_{t \to \infty} \left( -t^{-1} + 1 \right) = \underbrace{\lim_{t \to \infty} \left( -t^{-1}{} \right)}_{\text{=0}} + 1 = 0 + 1 = :success[1]

Note Summary

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Common Mistakes:

  1. Failing to recognise the need for limits: Improper integrals require setting up a limit if the range is infinite or the integrand is undefined.

  2. Incorrect integration bounds: When replacing the improper part with a parameter, ensure the bounds match the scenario (e.g., 0^+ for undefined values at 0).

  3. Forgetting to evaluate the limit: Always complete the problem by calculating the limit after integration.

  4. Misapplying the fundamental theorem of calculus: Ensure continuity of the function on the interval before applying integration.

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Key Formulas:

  1. Improper Integral with Infinite Bounds:
∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx\int_a^\infty f(x) \, dx = \lim_{t \to \infty} \int_a^t f(x) \, dx
  1. Improper Integral with Undefined Integrand:
∫abf(x) dx=lim⁡t→c±∫atf(x) dx\int_a^b f(x) \, dx = \lim_{t \to c^\pm} \int_a^t f(x) \, dx

where c is the point of discontinuity.

  1. Convergence of ∫1∞1xp dx\int_1^\infty \frac{1}{x^p} \, dx:
∫1∞1xp dx converges if and only if :highlight[p>1]\int_1^\infty \frac{1}{x^p} \, dx \text{ converges if and only if } :highlight[p > 1]
  1. Exponential Decay Integral:
∫0∞e−kx dx=1k,:highlight[k>0]\int_0^\infty e^{-kx} \, dx = \frac{1}{k}, \quad :highlight[k > 0]

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