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## Integral Calculus

### Course: Integral Calculus>Unit 1

Lesson 19: Improper integrals

# Improper integrals review

Review your knowledge of improper integrals.

## What are improper integrals?

Improper integrals are definite integrals that cover an unbounded area.
One type of improper integrals are integrals where at least one of the endpoints is extended to infinity. For example, integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x is an improper integral. It can be viewed as the limit limit, start subscript, b, \to, infinity, end subscript, integral, start subscript, 1, end subscript, start superscript, b, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x.
Another type of improper integrals are integrals whose endpoints are finite, but the integrated function is unbounded at one (or two) of the endpoints. For example, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x is an improper integral. It can be viewed as the limit limit, start subscript, a, \to, 0, start superscript, plus, end superscript, end subscript, integral, start subscript, a, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x.
An unbounded area that isn't infinite?! Is that for real?! Well, yeah! Not all improper integrals have a finite value, but some of them definitely do. When the limit exists we say the integral is convergent, and when it doesn't we say it's divergent.

## Practice set 1: Evaluating improper integrals with unbounded endpoints

Let's evaluate, for example, the improper integral integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x. As mentioned above, it's useful to view this integral as the limit limit, start subscript, b, \to, infinity, end subscript, integral, start subscript, 1, end subscript, start superscript, b, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x. We can use the fundamental theorem of calculus to find an expression for the integral:
\begin{aligned} \displaystyle\int_1^b\dfrac{1}{x^2}\,dx&=\displaystyle\int_1^b x^{-2}\,dx \\\\ &=\left[\dfrac{x^{-1}}{-1}\right]_1^b \\\\ &=\left[-\dfrac{1}{x}\right]_1^b \\\\ &=-\dfrac{1}{b}-\left(-\dfrac{1}{1}\right) \\\\ &=1-\dfrac{1}{b} \end{aligned}
Now we got rid of the integral and we have a limit to find:
\begin{aligned} \displaystyle\lim_{b\to\infty}\int_1^b\dfrac{1}{x^2}\,dx&=\displaystyle\lim_{b\to\infty}\left(1-\dfrac{1}{b}\right) \\\\ &=1-0 \\\\ &=1 \end{aligned}
Problem 1.1
• Current
integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, cubed, end fraction, d, x, equals, question mark

Want to try more problems like this? Check out this exercise.

## Practice set 2: Evaluating improper integrals with unbounded function

Let's evaluate, for example, the improper integral integral, start subscript, 0, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x. As mentioned above, it's useful to view this integral as the limit limit, start subscript, a, \to, 0, end subscript, integral, start subscript, a, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x. Again, we use the fundamental theorem of calculus to find an expression for the integral:
\begin{aligned} \displaystyle\int_a^1\dfrac{1}{\sqrt x}\,dx&=\displaystyle\int_a^1 x^{^{\large -\frac{1}{2}}}\,dx \\\\ &=\left[\dfrac{x^{^{\large\frac{1}{2}}}}{\frac{1}{2}}\right]_a^1 \\\\ &=\Bigl[2\sqrt x\Bigr]_a^1 \\\\ &=2\sqrt 1-2\sqrt a \\\\ &=2-2\sqrt a \end{aligned}
Now we got rid of the integral and we have a limit to find:
\begin{aligned} \displaystyle\lim_{a\to 0}\int_a^1\dfrac{1}{\sqrt x}\,dx&=\displaystyle\lim_{a\to 0}(2-2\sqrt a) \\\\ &=2-2\cdot 0 \\\\ &=2 \end{aligned}
Problem 2.1
• Current
integral, start subscript, 0, end subscript, start superscript, 8, end superscript, start fraction, 1, divided by, cube root of, x, end cube root, end fraction, d, x, equals, question mark