Questions: Math 172: Integration July 30, 2020 1 "Area under the curve" Integration is often introduced as a way to solve the problem of calculating areas of objects with curved boundaries. That problem is then set up as a problem involving the graph of a function. So integration becomes an operation on functions. It turns out to have many more uses than expressing "area under the curve". Here is a sketch of how integration is typically defined along these lines. Suppose f is a real-valued function with domain containing a closed interval [a, b]. We will determine a way to find the area between the graph of f and the x-axis, from x=a to x=b : We'll start by approximating this area as a sum of areas of rectangles (somewhat like the picture suggests). Let's start with an example. Say f(x)=x^2. Say we want to find the area of the striped region (from x=1 to x=3, above the x-axis, and below the graph):

Math 172: Integration

July 30, 2020

1 "Area under the curve" Integration is often introduced as a way to solve the problem of calculating areas of objects with curved boundaries. That problem is then set up as a problem involving the graph of a function. So integration becomes an operation on functions. It turns out to have many more uses than expressing "area under the curve".

Here is a sketch of how integration is typically defined along these lines. Suppose f is a real-valued function with domain containing a closed interval [a, b]. We will determine a way to find the area between the graph of f and the x-axis, from x=a to x=b :

We'll start by approximating this area as a sum of areas of rectangles (somewhat like the picture suggests).

Let's start with an example. Say f(x)=x^2. Say we want to find the area of the striped region (from x=1 to x=3, above the x-axis, and below the graph):
Transcript text: Math 172: Integration July 30, 2020 1 "Area under the curve" Integration is often introduced as a way to solve the problem of calculating areas of objects with curved boundaries. That problem is then set up as a problem involving the graph of a function. So integration becomes an operation on functions. It turns out to have many more uses than expressing "area under the curve". Here is a sketch of how integration is typically defined along these lines. Suppose $f$ is a real-valued function with domain containing a closed interval $[a, b]$. We will determine a way to find the area between the graph of $f$ and the $x$-axis, from $x=a$ to $x=b$ : We'll start by approximating this area as a sum of areas of rectangles (somewhat like the picture suggests). Let's start with an example. Say $f(x)=x^{2}$. Say we want to find the area of the striped region (from $x=1$ to $x=3$, above the $x$-axis, and below the graph): 1
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Solution

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Solution Steps

Step 1: Define the Function and Interval

We are given the function \( f(x) = x^2 \) and need to find the area under the curve from \( x = 1 \) to \( x = 3 \).

Step 2: Set Up the Integral

To find the area under the curve, we set up the definite integral of \( f(x) \) from 1 to 3: \[ \int_{1}^{3} x^2 \, dx \]

Step 3: Integrate the Function

Integrate \( x^2 \): \[ \int x^2 \, dx = \frac{x^3}{3} + C \]

Step 4: Evaluate the Definite Integral

Evaluate the integral from 1 to 3: \[ \left[ \frac{x^3}{3} \right]_{1}^{3} = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \]

Final Answer

The area under the curve \( f(x) = x^2 \) from \( x = 1 \) to \( x = 3 \) is \( \frac{26}{3} \).

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