Math / Calculus

Question11. Calculate $\int_{0}^{10}(8-x) d x$ in two ways: a. As the limit $\lim _{N \rightarrow \infty} R_{N}$ b. By sketching the relevant signed area and using geometry

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of $8-x$ from 0 to 10, first using limits of Riemann sums and then using geometry. Watch out! The $8-x$ term can be negative, so we need to be careful with signs when using geometry.

STEP 2

1. Calculate the integral as the limit of a Riemann sum.
2. Calculate the integral using geometry.

STEP 3

Alright, let's rock this Riemann sum!
We're going to slice the interval from 0 to 10 into $N$ equal pieces.
Each piece will have a width of $\Delta x = \frac{\textbf{10} - \textbf{0}}{N} = \frac{10}{N}$ .
The right endpoint of the $i$ -th interval is $x_i = 0 + i \cdot \Delta x = \frac{10i}{N}$ .
Our function is $f(x) = 8-x$ , so $f(x_i) = 8 - \frac{10i}{N}$ .

STEP 4

The Riemann sum $R_N$ is given by: $R_N = \sum_{i=1}^{N} f(x_i) \Delta x = \sum_{i=1}^{N} \left(8 - \frac{10i}{N}\right) \cdot \frac{10}{N}.$

STEP 5

Let's expand this sum: $R_N = \sum_{i=1}^{N} \left(\frac{80}{N} - \frac{100i}{N^2}\right) = \frac{80}{N} \sum_{i=1}^{N} 1 - \frac{100}{N^2} \sum_{i=1}^{N} i.$ We know that $\sum_{i=1}^{N} 1 = N$ and $\sum_{i=1}^{N} i = \frac{N(N+1)}{2}$ . Substituting these formulas, we get: $R_N = \frac{80}{N} \cdot N - \frac{100}{N^2} \cdot \frac{N(N+1)}{2} = 80 - \frac{50(N+1)}{N} = 80 - 50 - \frac{50}{N} = 30 - \frac{50}{N}.$

STEP 6

Now, let's take the limit as $N$ approaches infinity: $\lim_{N \rightarrow \infty} R_N = \lim_{N \rightarrow \infty} \left(30 - \frac{50}{N}\right) = 30 - 0 = \textbf{30}.$

STEP 7

The graph of $y = 8-x$ is a straight line with a y-intercept of 8 and a slope of -1.
It crosses the x-axis at $x=8$ .

STEP 8

The area above the x-axis is a triangle with vertices $(0,0)$ , $(0,8)$ , and $(8,0)$ .
Its area is $\frac{1}{2} \cdot \textbf{base} \cdot \textbf{height} = \frac{1}{2} \cdot 8 \cdot 8 = 32$ .

STEP 9

The area below the x-axis is a triangle with vertices $(8,0)$ , $(10,0)$ , and $(10,-2)$ .
Its area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$ .
Since it's below the x-axis, we consider this area negative, so we have $-2$ .

STEP 10

The total signed area is $32 + (-2) = \textbf{30}$ .

STEP 11

Using both methods, the definite integral $\int_{0}^{10}(8-x) dx$ is 30.

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Question11. Calculate $\int_{0}^{10}(8-x) d x$ in two ways: a. As the limit $\lim _{N \rightarrow \infty} R_{N}$ b. By sketching the relevant signed area and using geometry

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of $8-x$ from 0 to 10, first using limits of Riemann sums and then using geometry. Watch out! The $8-x$ term can be negative, so we need to be careful with signs when using geometry.

STEP 2

1. Calculate the integral as the limit of a Riemann sum.
2. Calculate the integral using geometry.

STEP 3

STEP 4

The Riemann sum $R_N$ is given by: $R_N = \sum_{i=1}^{N} f(x_i) \Delta x = \sum_{i=1}^{N} \left(8 - \frac{10i}{N}\right) \cdot \frac{10}{N}.$

STEP 5

STEP 6

Now, let's take the limit as $N$ approaches infinity: $\lim_{N \rightarrow \infty} R_N = \lim_{N \rightarrow \infty} \left(30 - \frac{50}{N}\right) = 30 - 0 = \textbf{30}.$

STEP 7

The graph of $y = 8-x$ is a straight line with a y-intercept of 8 and a slope of -1.
It crosses the x-axis at $x=8$ .

STEP 8

The area above the x-axis is a triangle with vertices $(0,0)$ , $(0,8)$ , and $(8,0)$ .
Its area is $\frac{1}{2} \cdot \textbf{base} \cdot \textbf{height} = \frac{1}{2} \cdot 8 \cdot 8 = 32$ .

STEP 9

The area below the x-axis is a triangle with vertices $(8,0)$ , $(10,0)$ , and $(10,-2)$ .
Its area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$ .
Since it's below the x-axis, we consider this area negative, so we have $-2$ .

STEP 10

The total signed area is $32 + (-2) = \textbf{30}$ .

STEP 11

Using both methods, the definite integral $\int_{0}^{10}(8-x) dx$ is 30.

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Question11. Calculate ∫010(8−x)dx\int_{0}^{10}(8-x) d x∫010​(8−x)dx in two ways: a. As the limit lim⁡N→∞RN\lim _{N \rightarrow \infty} R_{N}limN→∞​RN​ b. By sketching the relevant signed area and using geometry

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of 8−x8-x8−x from 0 to 10, first using limits of Riemann sums and then using geometry. Watch out! The 8−x8-x8−x term can be negative, so we need to be careful with signs when using geometry.

STEP 2

1. Calculate the integral as the limit of a Riemann sum.2. Calculate the integral using geometry.

STEP 3

STEP 4

The Riemann sum RNR_NRN​ is given by: RN=∑i=1Nf(xi)Δx=∑i=1N(8−10iN)⋅10N.R_N = \sum_{i=1}^{N} f(x_i) \Delta x = \sum_{i=1}^{N} \left(8 - \frac{10i}{N}\right) \cdot \frac{10}{N}.RN​=i=1∑N​f(xi​)Δx=i=1∑N​(8−N10i​)⋅N10​.

STEP 5

STEP 6

Now, let's take the **limit** as NNN approaches infinity: lim⁡N→∞RN=lim⁡N→∞(30−50N)=30−0=30.\lim_{N \rightarrow \infty} R_N = \lim_{N \rightarrow \infty} \left(30 - \frac{50}{N}\right) = 30 - 0 = \textbf{30}.N→∞lim​RN​=N→∞lim​(30−N50​)=30−0=30.

STEP 7

The graph of y=8−xy = 8-xy=8−x is a straight line with a y-intercept of 8 and a slope of -1.It crosses the x-axis at x=8x=8x=8.

STEP 8

The area above the x-axis is a triangle with vertices (0,0)(0,0)(0,0), (0,8)(0,8)(0,8), and (8,0)(8,0)(8,0).Its area is 12⋅base⋅height=12⋅8⋅8=32\frac{1}{2} \cdot \textbf{base} \cdot \textbf{height} = \frac{1}{2} \cdot 8 \cdot 8 = 3221​⋅base⋅height=21​⋅8⋅8=32.

STEP 9

The area below the x-axis is a triangle with vertices (8,0)(8,0)(8,0), (10,0)(10,0)(10,0), and (10,−2)(10,-2)(10,−2).Its area is 12⋅2⋅2=2\frac{1}{2} \cdot 2 \cdot 2 = 221​⋅2⋅2=2.Since it's below the x-axis, we consider this area **negative**, so we have −2-2−2.

STEP 10

The **total signed area** is 32+(−2)=3032 + (-2) = \textbf{30}32+(−2)=30.

STEP 11

Using both methods, the definite integral ∫010(8−x)dx\int_{0}^{10}(8-x) dx∫010​(8−x)dx is **30**.

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Question11. Calculate $\int_{0}^{10}(8-x) d x$ in two ways: a. As the limit $\lim _{N \rightarrow \infty} R_{N}$ b. By sketching the relevant signed area and using geometry

What is this asking? We need to find the definite integral of $8-x$ from 0 to 10, first using limits of Riemann sums and then using geometry. Watch out! The $8-x$ term can be negative, so we need to be careful with signs when using geometry.

1. Calculate the integral as the limit of a Riemann sum.
2. Calculate the integral using geometry.

The Riemann sum $R_N$ is given by: $R_N = \sum_{i=1}^{N} f(x_i) \Delta x = \sum_{i=1}^{N} \left(8 - \frac{10i}{N}\right) \cdot \frac{10}{N}.$

Now, let's take the limit as $N$ approaches infinity: $\lim_{N \rightarrow \infty} R_N = \lim_{N \rightarrow \infty} \left(30 - \frac{50}{N}\right) = 30 - 0 = \textbf{30}.$

The graph of $y = 8-x$ is a straight line with a y-intercept of 8 and a slope of -1.
It crosses the x-axis at $x=8$ .

The area above the x-axis is a triangle with vertices $(0,0)$ , $(0,8)$ , and $(8,0)$ .
Its area is $\frac{1}{2} \cdot \textbf{base} \cdot \textbf{height} = \frac{1}{2} \cdot 8 \cdot 8 = 32$ .

The area below the x-axis is a triangle with vertices $(8,0)$ , $(10,0)$ , and $(10,-2)$ .
Its area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$ .
Since it's below the x-axis, we consider this area negative, so we have $-2$ .

The total signed area is $32 + (-2) = \textbf{30}$ .

Using both methods, the definite integral $\int_{0}^{10}(8-x) dx$ is 30.