"While dealing with classes and grades think to yourself 'Do I want to perform as well as I can?' If the answer is yes, start changing your mindset to a growth mindset. The first step in doing this is learn, learn, learn. Don't worry about looking good, and having the best grade, just learn. It's so much more important than to get good grades. The second step is to work with passion and dedication - effort is key. People with growth mindsets know that they have to work hard, and they enjoy it. They understand that effort is what ignites their ability and causes it to grow. The last rule is to embrace your mistakes and confront your deficiencies. They don't blame others for their failures, they take it and get help."

– Clara Mitchinson (one of my awesome Day1: Great Lakes students) summarizing Carol Dweck's video and an article about her work.

Sketch by hand the following together on the same set of axes. Be sure to clearly indicate which function is above the other on the intervals \((0,1)\) and \((1,\infty)\).

1. \(\displaystyle y = \frac{1}{x}\) and \(\displaystyle y = \frac{1}{x^2}\).

2. \(\displaystyle y = \frac{1}{x^2}\) and \(\displaystyle y = \frac{1}{x^{1/2}}\).

3. \(\displaystyle y = e^x\) and \(y = \ln(x)\).

1. Given \(\displaystyle \int_{-\infty}^0 e^x \, dx = 1\), what is \(\displaystyle \int_0^1 \ln(x) \, dx\)? Draw a picture.

2. Given \(\displaystyle \int_1^\infty \frac{dx}{x^2} = 1\), what is \(\displaystyle \int_0^1 \frac{dx}{\sqrt{x}}\)? Draw a picture.

1. What is the definition of \(\ln(x)\) in terms of an integral?

2. Using the definition of \(\ln(x)\) in terms of an integral, explain why \(\ln(1/2)\) is negative.

Let \(f\) be a continuous function. Improper integrals are defined as limits of proper definite integrals.

1. \(\displaystyle \int_a^\infty f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx\)

2. \(\displaystyle \int_{-\infty}^b f(x) \, dx = \lim_{a \to -\infty} \int_a^b f(x) \, dx\)

3. \(\displaystyle \int_{-\infty}^{\infty} f(x) \, dx = \lim_{a \to -\infty} \int_a^c f(x) \, dx + \lim_{b \to \infty} \int_c^b f(x) \, dx\)

\[\int_1^{\infty} \frac{dx}{x^2} = \lim_{b \to \infty} \int_1^b \frac{dx}{x^2} = \lim_{b \to \infty} \bigg\lbrack \frac{-1}{x} \bigg\rbrack_1^b\]
\[ = \lim_{b \to \infty} \bigg( 1-\frac{1}{b} \bigg) = 1.\]

\[\int_{-\infty}^{c=2.5} f(x) \, dx = \lim_{a \to -\infty} \int_a^{c=2.5} f(x) \, dx.\]

\[\int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{c=2.5} f(x) \, dx + \int_{c=2.5}^{\infty} f(x) \, dx\] \[= \lim_{a \to -\infty} \int_{a}^{c} f(x) \, dx + \lim_{b \to \infty} \int_{c}^{b} f(x) \, dx.\]

Definition: Let \(f\) be a function that is continuous on \((a, b \rbrack\) and has a vertical asymptote \(x = a\). Define \[\int_a^b f(x) \, dx = \lim_{t \to a^+} \int_t^b f(x) \, dx.\]

A similar definition can be made for a function continuous on \(\lbrack a, b)\) which has a vertical asymptote at \(x=b\).

\[\int_0^1 \frac{dx}{x} = \lim_{a \to 0^+} \int_a^1 \frac{dx}{x} = \lim_{a \to 0^+} \bigg\lbrack \ln x \bigg\rbrack_a^1 = \infty\]

Evaluate \(\displaystyle \int_{-\infty}^{\infty} x e^{-x^2} \, dx\). Explain why your answer makes sense.

The improper integral \(\int_{-\infty}^{\infty} x \, dx\) is (choose one:)

1. convergent since \(\int_{-\infty}^0 x \, dx + \int_0^\infty x \, dx = -\infty + \infty = 0\).

2. convergent since it equals \[\lim_{t \to \infty} \int_{-t}^t x \, dx = \lim_{t \to \infty} \bigg( \frac{t^2}{2} - \frac{(-t)^2}{2} \bigg) = 0.\]

3. divergent since \(\int_{-\infty}^0 x \, dx + \int_0^\infty x \, dx = -\infty + \infty\) is an indeterminate form.

What is the role of the center (split) point?

Is \(\displaystyle \int_{-5}^{5} \frac{x \, dx}{ \sqrt{25-x^2} }\) an improper integral? Without evaluating this integral, explain why its value could be \(0\). Is its value \(0\)?

Does \(\displaystyle \int_{\pi}^{\infty} \cos(x) \, dx\) converge to a unique value, diverge to \(\infty\), diverge to \(-\infty\), or is it just

Draw an animation for this improper integral. How would you interpret it?

1. Does each function on \(x \geq 1\) have a convergent integral, divergent integral, or is it impossible to tell?

2. Is the area between \(f(x)\) and \(g(x)\) for \(x \geq 1\) finite or infinite?

An improper integral \(\displaystyle \int_1^\infty \frac{1}{x^p} \, dx\) is divergent when \(0 < p \leq 1\) and convergent when \(p > 1\). The reciprocal function \(\frac{1}{x} = x^{-1}\) has a divergent integral and is the "dividing line" between convergent and divergent integrals. For integrals \(\displaystyle \int_0^1 \frac{1}{x^p} \, dx\), use that \(\frac{1}{x^p}\) and \(\frac{1}{x^{1/p}}\) are inverse functions and graphs to help you quickly determine when the integral is convergent or divergent.