Improper Integrals: Videos & Practice Problems

Improper integrals involve integrating functions with infinite bounds. When the upper bound is infinity, rewrite the integral using a limit as a variable approaches infinity. For a lower bound of negative infinity, use a limit as the variable approaches negative infinity. If both bounds are infinite, split the integral into two parts. An integral is convergent if the limit exists and yields a finite number; otherwise, it is divergent. Understanding these concepts is crucial for evaluating the area under curves in calculus.

Improper Integrals: Infinite Intervals

In calculus, improper integrals arise when one or both bounds of an integral extend to infinity. To evaluate these integrals, we must use limits to handle the infinite bounds appropriately. An improper integral can be defined as follows: if we want to integrate a function from a lower bound \( a \) to an upper bound \( \infty \), we rewrite it as the limit of the integral from \( a \) to \( t \) as \( t \) approaches infinity:

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

This approach allows us to calculate the area under the curve from \( a \) to \( t \) and then extend that area to infinity. Similarly, if the lower bound is negative infinity, we express it as:

\[ \int_{-\infty}^b f(x) \, dx = \lim_{t \to -\infty} \int_t^b f(x) \, dx \]

In both cases, we cannot directly substitute infinity into the integral; instead, we use a variable \( t \) that approaches infinity or negative infinity. When both bounds are infinite, we can split the integral into two parts, using a constant \( c \) between the bounds:

\[ \int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^c f(x) \, dx + \int_c^\infty f(x) \, dx \]

Next, we determine whether an improper integral is convergent or divergent. An integral is considered convergent if the limit exists and yields a finite number, while it is divergent if the limit does not exist or approaches infinity.

For example, consider the integral of \( e^x \) from \( 0 \) to \( \infty \):

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

The integral of \( e^x \) is \( e^x \), so we evaluate:

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

As \( t \) approaches infinity, \( e^t \) also approaches infinity, leading to a divergent integral.

In another example, consider the integral from \( -\infty \) to \( 0 \):

\[ \int_{-\infty}^0 e^x \, dx = \lim_{t \to -\infty} \int_t^0 e^x \, dx \]

Evaluating this gives:

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

Here, the integral converges to 1.

Finally, for the integral from \( -\infty \) to \( \infty \):

\[ \int_{-\infty}^\infty e^x \, dx = \int_{-\infty}^0 e^x \, dx + \int_0^\infty e^x \, dx \]

Since the first integral converges to 1 and the second diverges to infinity, the overall integral is divergent.

Understanding these concepts and procedures is crucial for effectively working with improper integrals in calculus. Practice with various examples will help solidify these techniques and enhance your problem-solving skills.

Improper Integrals: Infinite Intervals Example 1

In evaluating improper integrals, it is essential to determine whether they converge or diverge, and to compute their values if they converge. For instance, consider the integral of the function \( x e^x \) from 0 to infinity. Since the upper limit is infinity, we rewrite the integral as a limit:

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

To solve this integral, we apply integration by parts, which is based on the formula:

\[ \int u \, dv = uv - \int v \, du \]

Choosing \( u = x \) (thus \( du = dx \)) and \( dv = e^x \, dx \) (which gives \( v = e^x \)), we can express the integral as:

\[ \lim_{t \to \infty} \left( x e^x - \int e^x \, dx \right) \]

Evaluating the integral of \( e^x \) yields:

\[ \lim_{t \to \infty} \left( x e^x - e^x \right) \bigg|_0^t \]

Substituting the bounds, we find:

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

As \( t \) approaches infinity, both \( t e^t \) and \( e^t \) diverge to infinity, leading to the conclusion that the integral diverges. Thus, the result for this example is that the integral diverges.

In the second example, we evaluate the integral of \( x e^x \) from negative infinity to 0. This can be expressed as:

\[ \int_{-\infty}^0 x e^x \, dx = \lim_{t \to -\infty} \int_t^0 x e^x \, dx \]

Using the same integration by parts method as before, we find that the integral evaluates to:

\[ \lim_{t \to -\infty} \left( 0 - (t e^t - e^t) \right) \bigg|_t^0 \]

Evaluating this limit, we find that as \( t \) approaches negative infinity, \( t e^t \) approaches 0, leading to a final result of:

\[ 0 - 1 = -1 \]

Thus, this integral converges to -1.

Finally, for the integral from negative infinity to positive infinity of \( x e^x \), we can split it into two parts:

\[ \int_{-\infty}^\infty x e^x \, dx = \int_{-\infty}^0 x e^x \, dx + \int_0^\infty x e^x \, dx \]

From previous evaluations, we know that the first integral converges to -1, while the second diverges. Therefore, the overall integral diverges, as the divergence of one part dominates the result.

In summary, understanding the behavior of the function as it approaches infinity or negative infinity is crucial in determining the convergence or divergence of improper integrals. This process often involves techniques such as integration by parts and careful limit evaluation.

Evaluate the integral or state that it diverges. 
$\int_{-\infty}^{-1}\frac{2}{x^3}dx$

• A. The integral diverges.
• B. $-2$; converges.
• C. $-1$; converges.
• D. $1$; converges.

Answer: (C)

Evaluate the integral or state that it diverges. 
$\int_2^{\infty}\frac{1}{x(lnx)^4}dx$

• A. The Integral diverges.
• B. $2$; converges.
• C. $\frac{1}{5\left(\ln2\right)^5}$ ; converges.
• D. $\frac{1}{3\left(\ln2\right)^3}$; converges.

Answer: (D)

Here’s what students ask on this topic:

What is an improper integral and how is it different from a regular definite integral?

An improper integral is a type of definite integral where one or both of the bounds are infinite, or the integrand has an infinite discontinuity within the interval of integration. Unlike regular definite integrals, improper integrals require additional steps to evaluate. For infinite bounds, the integral is rewritten using a limit as a variable approaches infinity or negative infinity. For discontinuities, the interval is split at the point of discontinuity, and limits are applied to evaluate the integral. The result of an improper integral can be either convergent (finite value) or divergent (does not exist).

How do you evaluate an improper integral with an infinite upper bound?

To evaluate an improper integral with an infinite upper bound, rewrite the integral by replacing the upper bound with a variable, such as t, and then take the limit as t approaches infinity. For example, the integral $\int a_{\infty }f\left(x\right)dx$ becomes $\int a_{t}f\left(x\right)dx$ with $\lim (t\to \infty )$. Solve the integral as usual, then apply the limit to determine if the result is finite (convergent) or infinite (divergent).

What does it mean for an improper integral to be convergent or divergent?

An improper integral is convergent if the limit used to evaluate the integral exists and results in a finite value. This means the area under the curve is finite despite infinite bounds or discontinuities. Conversely, an improper integral is divergent if the limit does not exist or results in infinity, indicating the area under the curve is infinite or undefined. For example, the integral $\int 0_{\infty }{e}^x$ is divergent because the exponential function grows infinitely large as x approaches infinity.

How do you handle improper integrals with both bounds being infinite?

When both bounds of an improper integral are infinite, split the integral into two parts at a constant c. For example, the integral $\int {-}_{\infty }\infty f\left(x\right)dx$ becomes $\int {-}_{\infty }cf\left(x\right)dx$ + $\int c_{\infty }f\left(x\right)dx$. Evaluate each part using limits for the infinite bounds. If either part is divergent, the entire integral is divergent.

What are the steps to determine if an improper integral is convergent or divergent?

To determine if an improper integral is convergent or divergent, follow these steps:
1. Identify if the integral has infinite bounds or discontinuities.
2. Rewrite the integral using limits for infinite bounds or split the interval at the discontinuity.
3. Solve the integral as usual.
4. Apply the limit to evaluate the result.
5. If the limit exists and yields a finite value, the integral is convergent. If the limit does not exist or results in infinity, the integral is divergent. For example, $\int 0_{\infty }{e}^{-}$ is convergent because the limit results in a finite value.

Why can’t infinity be directly used as a bound in improper integrals?

Infinity cannot be directly used as a bound in improper integrals because it is not a real number. Integrals require numerical bounds to calculate the area under a curve. Instead, infinity is approached using limits. For example, the integral $\int a_{\infty }f\left(x\right)dx$ is rewritten as $\lim (t\to \infty )\int a_{t}f\left(x\right)dx$. This approach ensures the integral is mathematically valid.