diff --git a/course/.mapping.json b/course/.mapping.json index 53cdfa5..d0c1a7a 100755 --- a/course/.mapping.json +++ b/course/.mapping.json @@ -1 +1 @@ -{"source/main.ptx": ["frontmatter", "course-documents", "ch-notes", "quizzes", "recitations"], "source/syllabus/syllabus.ptx": ["syllabus"], "source/notes/1-13.ptx": ["notes-01-13"], "source/notes/1-15.ptx": ["notes-01-15"], "source/notes/1-20.ptx": ["notes-01-20"], "source/notes/1-22.ptx": ["notes-01-22"], "source/notes/1-27.ptx": ["notes-01-27"], "source/quizzes/quiz-01.ptx": ["quiz-01"], "source/recitations/calculus-review.ptx": ["recitation-calculus-review"]} \ No newline at end of file +{"source/main.ptx": ["frontmatter", "course-documents", "ch-notes", "quizzes", "recitations"], "source/syllabus/syllabus.ptx": ["syllabus"], "source/notes/1-13.ptx": ["notes-01-13"], "source/notes/1-15.ptx": ["notes-01-15"], "source/notes/1-20.ptx": ["notes-01-20"], "source/notes/1-22.ptx": ["notes-01-22"], "source/notes/1-27.ptx": ["notes-01-27"], "source/notes/1-29.ptx": ["notes-01-29"], "source/quizzes/quiz-01.ptx": ["quiz-01"], "source/recitations/calculus-review.ptx": ["recitation-calculus-review"]} \ No newline at end of file diff --git a/course/ch-notes.html b/course/ch-notes.html index 2385434..c180b8b 100755 --- a/course/ch-notes.html +++ b/course/ch-notes.html @@ -211,6 +211,10 @@ eBookConfig.enable_chatcodes = false;
  • Joint Distributions
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    Thursday, Jan 29
    + +
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  • Tuesday, Jan 20
  • Thursday, Jan 22
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    • +
  • Thursday, Jan 29
    • πŸ”—
    PrevTopNext diff --git a/course/course-documents.html b/course/course-documents.html index 068a220..5dbf87d 100755 --- a/course/course-documents.html +++ b/course/course-documents.html @@ -211,6 +211,10 @@ eBookConfig.enable_chatcodes = false;
  • Joint Distributions
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    Thursday, Jan 29
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  • diff --git a/course/frontmatter.html b/course/frontmatter.html index 78284b9..7560115 100755 --- a/course/frontmatter.html +++ b/course/frontmatter.html @@ -211,6 +211,10 @@ eBookConfig.enable_chatcodes = false;
  • Joint Distributions
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    Thursday, Jan 29
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  • diff --git a/course/index.html b/course/index.html index 37a69e0..d4bb52f 100755 --- a/course/index.html +++ b/course/index.html @@ -2,7 +2,7 @@ - + diff --git a/course/knowl/xref/def-discrete-EV.html b/course/knowl/xref/def-discrete-EV.html new file mode 100755 index 0000000..f6103d8 --- /dev/null +++ b/course/knowl/xref/def-discrete-EV.html @@ -0,0 +1,35 @@ + + + + + + + + + + + + + + + + + + + + +

    +Definition 73. +

    +
    +
    Let \(X\) be a discrete random variable taking values \(x_1, x_2, \dotsc, x_n\) with probabilities \(p_1, p_2, \dotsc, p_n\text{.}\) The expected value of \(X\) is:
    +
    +\begin{gather*} +\E(X) = \sum_{i=1}^n x_ip_i = x_1p_1 + x_2p_2 + \dotsb + x_np_n. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—
    in-context + + diff --git a/course/lunr-pretext-search-index.js b/course/lunr-pretext-search-index.js index acd55fe..6e56c4f 100755 --- a/course/lunr-pretext-search-index.js +++ b/course/lunr-pretext-search-index.js @@ -612,6 +612,159 @@ var ptx_lunr_docs = [ "title": "", "body": " Let indicate heads on the first and second flip, respectively, of a fair coin. Then: Joint Distribution for Indicator Random Variables 1\/4 1\/4 1\/4 1\/4 Then the marginal distributions are: Then, for any : So are independent. " }, +{ + "id": "notes-01-29", + "level": "1", + "url": "notes-01-29.html", + "type": "Section", + "number": "", + "title": "Thursday, Jan 29", + "body": " Thursday, Jan 29 This is an outline of the topics we covered in class. These notes are not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes. Expected Value Let be the roll of a fair D6. What is the average value of ? The sample space is . To find the average: That is: take each value of , multiply by the probability, and add all the results together. Given a distribution table: Distribution 1 0.1 2 0.1 3 0.1 4 0.1 5 0.1 6 0.5 Then the average is: Let be a discrete random variable taking values with probabilities . The expected value of is: Let indicate , with . Then: That is: \"E(indicator random variable) = Pr(event that it indicates)\". Suppose we flip a fair coin 100 times. Let indicate heads on each flip. Then , so for every . What if indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails. Since these flip results are independent, the probabilities multiply: Let be a continuous random variable with pdf . The expected value of is: It's worth putting this side-by-side with to compare the structure of each formula. These both say: \"multiply each value of the random variable by the probability, then accumulate all of those products\". Expected value is a weighted average of random variable values, with the probabilities as the weights. Let with pdf . Then: What would happen if you forgot the ? Then the calculation would become: This mistake will often be easy to catch. For example, this random variable takes values between 0 and 1, so it doesn't seem very likely that the average value is 1! Find given a distribution for . Distribution for -1 0.4 1 0.1 2 0.2 3 0.3 We can start by writing a distribution table for : Distribution for 1 0.5 4 0.2 9 0.3 Then: If takes on values with probabilities , and is any function, then: Revisiting the previous example, the theorem says we don't need to first create the distribution table for . We can use the distribution table for , and just apply the square to each value: The situation with continuous random variables is similar. Let with pdf . Then: Notice, in particular, that If are random variables with finite expected value and , then: Let be the number of heads in coin flips with bias . Then . Gross. Instead of calculating this directly, define to indicate heads on each flip. Then: Flip a fair coin 100 times. Then, the expected number of heads is: Flip a fair coin 100 times. What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables each indicating a run of 4 heads starting at the specified flip. Then for each . Let be the number of runs of 4 heads. Then: Consider a geometric distribution . Geometric Distribution with 1 2 3 4 Then is an infinite summation. Instead, consider the following argument. If we flip a coin until we see heads, we either see heads on flip 1 or not. In the first case, . In the second case, what is the average value of ? Starting at flip 2, it will take on average flips to see heads. Since we already flipped the coin once, the total number of flips will be . So, we can write: " +}, +{ + "id": "notes-01-29-3-2", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-2", + "type": "Example", + "number": "70", + "title": "", + "body": " Let be the roll of a fair D6. What is the average value of ? The sample space is . To find the average: That is: take each value of , multiply by the probability, and add all the results together. " +}, +{ + "id": "notes-01-29-3-3", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-3", + "type": "Example", + "number": "71", + "title": "", + "body": " Given a distribution table: Distribution 1 0.1 2 0.1 3 0.1 4 0.1 5 0.1 6 0.5 Then the average is: " +}, +{ + "id": "def-discrete-EV", + "level": "2", + "url": "notes-01-29.html#def-discrete-EV", + "type": "Definition", + "number": "73", + "title": "", + "body": " Let be a discrete random variable taking values with probabilities . The expected value of is: " +}, +{ + "id": "notes-01-29-3-5", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-5", + "type": "Example", + "number": "74", + "title": "", + "body": " Let indicate , with . Then: That is: \"E(indicator random variable) = Pr(event that it indicates)\". " +}, +{ + "id": "notes-01-29-3-6", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-6", + "type": "Example", + "number": "75", + "title": "", + "body": " Suppose we flip a fair coin 100 times. Let indicate heads on each flip. Then , so for every . What if indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails. Since these flip results are independent, the probabilities multiply: " +}, +{ + "id": "def-continuous-EV", + "level": "2", + "url": "notes-01-29.html#def-continuous-EV", + "type": "Definition", + "number": "76", + "title": "", + "body": " Let be a continuous random variable with pdf . The expected value of is: " +}, +{ + "id": "notes-01-29-3-9", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-9", + "type": "Example", + "number": "77", + "title": "", + "body": " Let with pdf . Then: What would happen if you forgot the ? Then the calculation would become: This mistake will often be easy to catch. For example, this random variable takes values between 0 and 1, so it doesn't seem very likely that the average value is 1! " +}, +{ + "id": "notes-01-29-3-10", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-10", + "type": "Example", + "number": "78", + "title": "", + "body": " Find given a distribution for . Distribution for -1 0.4 1 0.1 2 0.2 3 0.3 We can start by writing a distribution table for : Distribution for 1 0.5 4 0.2 9 0.3 Then: " +}, +{ + "id": "notes-01-29-3-11", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-11", + "type": "Theorem", + "number": "81", + "title": "", + "body": " If takes on values with probabilities , and is any function, then: " +}, +{ + "id": "notes-01-29-3-12", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-12", + "type": "Example", + "number": "82", + "title": "", + "body": " Revisiting the previous example, the theorem says we don't need to first create the distribution table for . We can use the distribution table for , and just apply the square to each value: " +}, +{ + "id": "notes-01-29-3-13", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-13", + "type": "Example", + "number": "83", + "title": "", + "body": " The situation with continuous random variables is similar. Let with pdf . Then: Notice, in particular, that " +}, +{ + "id": "thm-linearity", + "level": "2", + "url": "notes-01-29.html#thm-linearity", + "type": "Theorem", + "number": "84", + "title": "", + "body": " If are random variables with finite expected value and , then: " +}, +{ + "id": "notes-01-29-3-15", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-15", + "type": "Example", + "number": "85", + "title": "", + "body": " Let be the number of heads in coin flips with bias . Then . Gross. Instead of calculating this directly, define to indicate heads on each flip. Then: " +}, +{ + "id": "notes-01-29-3-16", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-16", + "type": "Example", + "number": "86", + "title": "", + "body": " Flip a fair coin 100 times. Then, the expected number of heads is: " +}, +{ + "id": "notes-01-29-3-17", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-17", + "type": "Example", + "number": "87", + "title": "", + "body": " Flip a fair coin 100 times. What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables each indicating a run of 4 heads starting at the specified flip. Then for each . Let be the number of runs of 4 heads. Then: " +}, +{ + "id": "notes-01-29-3-18", + "level": "2", + "url": "notes-01-29.html#notes-01-29-3-18", + "type": "Example", + "number": "88", + "title": "", + "body": " Consider a geometric distribution . Geometric Distribution with 1 2 3 4 Then is an infinite summation. Instead, consider the following argument. If we flip a coin until we see heads, we either see heads on flip 1 or not. In the first case, . In the second case, what is the average value of ? Starting at flip 2, it will take on average flips to see heads. Since we already flipped the coin once, the total number of flips will be . So, we can write: " +}, { "id": "quiz-01", "level": "1", @@ -698,7 +851,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "recitation-calculus-review.html#thm-FTC", "type": "Theorem", - "number": "70", + "number": "90", "title": "Fundamental Theorem of Calculus.", "body": " Fundamental Theorem of Calculus If is an antiderivative of i.e., if then: " }, diff --git a/course/notes-01-13.html b/course/notes-01-13.html index 5e7e209..08a6348 100755 --- a/course/notes-01-13.html +++ b/course/notes-01-13.html @@ -211,6 +211,10 @@ eBookConfig.enable_chatcodes = false;
  • Joint Distributions
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    Thursday, Jan 29
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  • Joint Distributions
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  • Joint Distributions
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    diff --git a/course/notes-01-29.html b/course/notes-01-29.html new file mode 100755 index 0000000..0766358 --- /dev/null +++ b/course/notes-01-29.html @@ -0,0 +1,588 @@ + + + + + + + + + + + + + + + + +Thursday, Jan 29 + + + + + + + + + + + + + + + + + + + + +Skip to main content
    +

    Introduction to Probability and Statistics for the Life Sciences Spring 2026 MATH 1044

    +

    Andy Eisenberg

    +
    +
    + +

    +Section Thursday, Jan 29 +

    +
    This is an outline of the topics we covered in class. These notes are not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
    πŸ”—
    +

    +Subsection Expected Value +

    +

    +Example 70. +

    +
    +
    Let \(R\) be the roll of a fair D6. What is the average value of \(R\text{?}\) The sample space is \(\Omega = \{1, 2, 3, 4, 5, 6\}\text{.}\) To find the average:
    +
    +\begin{align*} +\text{Avg} \amp = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right) \\ +\amp = \frac{21}{6} = \frac{7}{2} = 3.5 +\end{align*} +
    +
    That is: take each value of \(R\text{,}\) multiply by the probability, and add all the results together.
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 71. +

    +
    Given a distribution table:
    πŸ”—
    +
    Table 72. Distribution
    πŸ”—
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    \(s\)\(\Pr(S = s)\)
    10.1
    20.1
    30.1
    40.1
    50.1
    60.5
    +
    Then the average is:
    +
    +\begin{gather*} +\text{Avg} = 1(0.1) + 2(0.1) + 3(0.1) + 4(0.1) + 5(0.1) + 6(0.5) = 4.5. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Definition 73. +

    +
    +
    Let \(X\) be a discrete random variable taking values \(x_1, x_2, \dotsc, x_n\) with probabilities \(p_1, p_2, \dotsc, p_n\text{.}\) The expected value of \(X\) is:
    +
    +\begin{gather*} +\E(X) = \sum_{i=1}^n x_ip_i = x_1p_1 + x_2p_2 + \dotsb + x_np_n. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 74. +

    +
    +
    Let \(X\) indicate \(A\text{,}\) with \(\Pr(A) = p\text{.}\) Then:
    +
    +\begin{gather*} +E(X) = 0 \cdot \underbrace{\Pr(X = 0)}_{1 - p} + 1 \cdot \underbrace{\Pr(X = 1)}_{p} = 0(1-p) + 1p = p. +\end{gather*} +
    +
    That is: "E(indicator random variable) = Pr(event that it indicates)".
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 75. +

    +
    Suppose we flip a fair coin 100 times. Let \(H_1, H_2, \dotsc, H_{100}\) indicate heads on each flip. Then \(\Pr(H_i = 1) = \frac{1}{2}\text{,}\) so \(E(H_i) = \frac{1}{2}\) for every \(i\text{.}\)
    πŸ”—
    +
    +
    What if \(X\) indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails. Since these flip results are independent, the probabilities multiply:
    +
    +\begin{gather*} +\Pr(X = 1) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{16} = \E(X). +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Definition 76. +

    +
    +
    Let \(X \in [a, b]\) be a continuous random variable with pdf \(f(x)\text{.}\) The expected value of \(X\) is:
    +
    +\begin{gather*} +\E(X) = \int_a^b x f(x)\ dx. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—
    It’s worth putting this side-by-side with DefinitionΒ 73 to compare the structure of each formula. These both say: "multiply each value of the random variable by the probability, then accumulate all of those products". Expected value is a weighted average of random variable values, with the probabilities as the weights.
    πŸ”—
    +
    +

    +Example 77. +

    +
    +
    Let \(X \in [0, 1]\) with pdf \(f(x) = \frac{3}{2}\sqrt{x}\text{.}\) Then:
    +
    +\begin{align*} +\E(X) \amp = \int_0^1 x f(x)\ dx \\ +\amp = \int_0^1 x \frac{3}{2} x^{1/2}\ dx \\ +\amp = \int_0^1 \frac{3}{2} x^{3/2}\ dx \\ +\amp = \frac{3}{2} \frac{2x^{5/2}}{5}\bigg|_0^1 \\ +\amp = \frac{3}{5} \cdot 1^{5/2} - \frac{3}{5} \cdot 0^{5/2} \\ +\amp = \frac{3}{5}. +\end{align*} +
    +
    What would happen if you forgot the \(x\text{?}\) Then the calculation would become:
    +
    +\begin{align*} +\int_0^1 f(x)\ dx = 1 = \text{total probability} \amp +\end{align*} +
    +
    This mistake will often be easy to catch. For example, this random variable takes values between 0 and 1, so it doesn’t seem very likely that the average value is 1!
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 78. +

    +
    Find \(\E(X^2)\) given a distribution for \(X\text{.}\)
    πŸ”—
    +
    Table 79. Distribution for \(X\)
    πŸ”—
    + + + + + + + + + + + + + + + + + + + + +
    \(x\)\(\Pr(X = x)\)
    -10.4
    10.1
    20.2
    30.3
    We can start by writing a distribution table for \(X^2\text{:}\)
    πŸ”—
    +
    Table 80. Distribution for \(X^2\)
    πŸ”—
    + + + + + + + + + + + + + + + + +
    \(x\)\(\Pr(X^2 = x)\)
    10.5
    40.2
    90.3
    +
    Then:
    +
    +\begin{gather*} +\E(X^2) 1(0.5) + 4(0.2) + 9(0.3) = 4. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 82. +

    +
    +
    Revisiting the previous example, the theorem says we don’t need to first create the distribution table for \(X^2\text{.}\) We can use the distribution table for \(X\text{,}\) and just apply the square to each value:
    +
    +\begin{gather*} +\E(X^2) = (-1)^2(0.4) + (1)^2(0.1) + (2)^2(0.2) + (3)^2(0.3) = 4. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 83. +

    +
    +
    The situation with continuous random variables is similar. Let \(X\in [0, 1]\) with pdf \(f(x) = \frac{3}{2}\sqrt{x}\text{.}\) Then:
    +
    +\begin{align*} +\E(X^2) \amp = \int_0^1 x^2 f(x)\ dx \\ +\amp = \int_0^1 x^2 \frac{3}{2} x^{1/2}\ dx \\ +\amp = \int_0^1 \frac{3}{2} x^{5/2}\ dx \\ +\amp = \frac{3}{2}\cdot \frac{2x^{7/2}}{7}\bigg|_0^1 \\ +\amp = \frac{3}{7}\cdot 1^{7/2} - 0\cdot 0^{7/2} \\ +\amp = \frac{3}{7}. +\end{align*} +
    +
    Notice, in particular, that \(\E(X^2) \neq \E(X).\) +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 85. +

    +
    +
    Let \(N\) be the number of heads in \(n\) coin flips with bias \(p\text{.}\) Then \(N\sim \Bin(n, p)\text{.}\) +
    +
    +\begin{gather*} +\E)N) = \sum_{i=0}^n i\cdot b(i; n, p) = \sum_{i = 0}^n i \cdot {n \choose i} p^i (1-p)^{n - i}. +\end{gather*} +
    +
    Gross.
    +
    πŸ”—
    +
    +
    Instead of calculating this directly, define \(H_1, H_2, \dotsc, H_n\) to indicate heads on each flip. Then:
    +
    +\begin{align*} +N \amp = H_1 + H_2 + \dotsb + H_n \\ +\E(N) \amp = \E(H_1 + H_2 + \dotsb + H_n) \\ +\E(N) \amp = \E(H_1) + \E(H_2) + \dotsb + \E(H_n) \\ +\amp = p + p + \dotsb + p \\ +\amp = np. +\end{align*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 86. +

    +
    +
    Flip a fair coin 100 times. Then, the expected number of heads is:
    +
    +\begin{gather*} +\underbrace{(100)}_{n}\underbrace{(0.5)}_{p} = 50. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 87. +

    +
    +
    Flip a fair coin 100 times. What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables \(R_1, R_2, \dotsc, R_{97}\) each indicating a run of 4 heads starting at the specified flip. Then \(\E(R_i) = \frac{1}{16}\) for each \(i\text{.}\) Let \(T\) be the number of runs of 4 heads. Then:
    +
    +\begin{align*} +\E(T) \amp = \E(R_1 + \dotsb + R_{97}) \\ +\amp = \E(R_1) + \dotsb + (R_{97}) \\ +\amp = \frac{1}{16} + \dotsb + \frac{1}{16} \\ +\amp = \frac{97}{16} \approx 6. +\end{align*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 88. +

    +
    Consider a geometric distribution \(T \sim \Geom(p)\text{.}\)
    πŸ”—
    +
    Table 89. Geometric Distribution
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    \(k\)\(\Pr(T = k)\)with \(p = \frac{1}{2}\) +
    1\(p\)\(\frac{1}{2}\)
    2\((1-p)p\)\(\frac{1}{4}\)
    3\((1-p)^2p\)\(\frac{1}{8}\)
    4\((1-p)^3p\)\(\frac{1}{16}\)
    \(\vdots\)\(\vdots\)\(\vdots\)
    Then \(\E(T) = (1)\left(\frac{1}{2}\right) + (2)\left(\frac{1}{4}\right) + (3)\left(\frac{1}{8}\right) + \dotsb\) is an infinite summation.
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    +
    +
    Instead, consider the following argument. If we flip a coin until we see heads, we either see heads on flip 1 or not. In the first case, \(T = 1\text{.}\) In the second case, what is the average value of \(T\text{?}\) Starting at flip 2, it will take on average \(\E(T)\) flips to see heads. Since we already flipped the coin once, the total number of flips will be \(1 + \E(T)\text{.}\) So, we can write:
    +
    +\begin{align*} +\E(T) \amp = 1 \left(\frac{1}{2}\right) + (1 + \E(T)) \left(1 - \frac{1}{2}\right) \\ +\E(T) \amp = \frac{1}{2} + \frac{1}{2} + \frac{\E(T)}{2} \\ +\E(T) - \frac{\E(T)}{2} \amp = 1 \\ +\frac{\E(T)}{2} \amp = 1 \\ +\E(T) \amp = 2. +\end{align*} +
    +
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    • -PrevUpNext +PrevUpNext
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    • πŸ”—
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    -Theorem 70. Fundamental Theorem of Calculus. +Theorem 90. Fundamental Theorem of Calculus.

    If \(F(x)\) is an antiderivative of \(f(x)\)β€”i.e., if \(F'(x) = f(x)\)β€”then:
    @@ -299,7 +303,7 @@ eBookConfig.enable_chatcodes = false;
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    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
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    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
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    5.

    Practice with the following integrals:
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    • πŸ”—
    πŸ”—

    -Theorem 70. Fundamental Theorem of Calculus. +Theorem 90. Fundamental Theorem of Calculus.

    If \(F(x)\) is an antiderivative of \(f(x)\)β€”i.e., if \(F'(x) = f(x)\)β€”then:
    @@ -284,7 +288,7 @@ eBookConfig.enable_chatcodes = false;
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    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
    πŸ”—
    +
    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
    πŸ”—

    5.

    Practice with the following integrals:
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