5.2.1 Improper Integrals

What Are Improper Integrals?

An improper integra****l arises when:

The integrand is undefined at one or more points within the range of integration.
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:

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

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

Solution Method:

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

\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:

\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:

\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 $\int_1^\infty \frac{1}{x^2} \, dx$

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

Replace the infinite limit with a parameter $t$ :

\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 $\frac{1}{x^2}$ is $\frac{1}{x}$ :

\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 $t \to \infty$ , $\frac{1}{t} \to 0$ :

\lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 0 + 1 = 1

Result:

\int_1^\infty \frac{1}{x^2} \, dx = :success[1]

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Example 2: Evaluate $\int_0^2 \frac{1}{\sqrt{x}} \, dx$

Step 1: Set up the limit:

The integrand is undefined at x = 0.

Replace this limit with $t$ :

\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 $\frac{1}{\sqrt{x}}$ is $2\sqrt{x}$ :

\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 $t \to 0^+$ , $\sqrt{t} \to 0$

\lim_{t \to 0^+} \left( 2\sqrt{2} - 2\sqrt{t} \right) = 2\sqrt{2}

Result:

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

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Example 3: Evaluate $\int_0^\infty e^{-x} \, dx$

Step 1: Set up the limit:

The upper limit is infinite:

\int_0^\infty e^{-x} \, dx = \lim_{t \to \infty} \int_0^t e^{-x} \, dx

Step 2: Integrate:

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

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

Step 3: Take the limit:

As $t \to \infty$ , $e^{-t} \to 0$ :

\lim_{t \to \infty} \left( -e^{-t} + 1 \right) = 0 + 1 = 1

Result:

\int_0^\infty e^{-x} \, dx = :success[1]

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Example 4:

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

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

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

Step 2: Show the process of $t$ approaching the undefined limit:

\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:

Failing to recognize the need for limits: Improper integrals require setting up a limit if the range is infinite or the integrand is undefined.
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).
Forgetting to evaluate the limit: Always complete the problem by calculating the limit after integration.
Misapplying the fundamental theorem of calculus: Ensure continuity of the function on the interval before applying integration.

infoNote

Key Formulas:

Improper Integral with Infinite Bounds:

\int_a^\infty f(x) \, dx = \lim_{t \to \infty} \int_a^t f(x) \, dx

Improper Integral with Undefined Integrand:

\int_a^b f(x) \, dx = \lim_{t \to c^\pm} \int_a^t f(x) \, dx

where c is the point of discontinuity.

Convergence of $\int_1^\infty \frac{1}{x^p} \, dx$ :

\int_1^\infty \frac{1}{x^p} \, dx \text{ converges if and only if } :highlight[p > 1]

Exponential Decay Integral:

\int_0^\infty e^{-kx} \, dx = \frac{1}{k}, \quad :highlight[k > 0]

Improper Integrals Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

5.2.1 Improper Integrals

What Are Improper Integrals?

How to Evaluate Improper Integrals

Undefined Integrand at a Point (Type I):

Infinite Limits of Integration (Type II):

Worked Examples

Note Summary

Common Mistakes:

Key Formulas:

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