Math / Calculus

QuestionCompute the average of $f(x)=x^{2}+x+1$ over the interval $[0,1]$ . If your answer is not a whole number, round accurate to at least two decimal places.
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Studdy Solution

STEP 1

What is this asking? Find the average value of a parabola-shaped function between two points. Watch out! Don't just plug in the endpoints!
We need the average value of the function across the entire interval.

STEP 2

1. Define the function and interval
2. Set up the average value formula
3. Evaluate the integral
4. Calculate the result

STEP 3

Alright, so we've got our function $f(x) = x^2 + x + 1$ , which makes a lovely U-shaped parabola!
We're looking at it between $x = \mathbf{0}$ and $x = \mathbf{1}$ , so that's our interval!

STEP 4

The average value of a function over an interval $[a, b]$ is given by the formula: $\frac{1}{b - a} \cdot \int_{a}^{b} f(x) \, dx$ This formula basically says, "find the area under the curve between a and b, and then divide by the width of the interval".
It's like finding the height of a rectangle with the same area and width!

STEP 5

In our case, $a = \mathbf{0}$ and $b = \mathbf{1}$ , and our function is $f(x) = x^2 + x + 1$ .
So, plugging those values into our formula, we get: $\frac{1}{1 - 0} \cdot \int_{0}^{1} (x^2 + x + 1) \, dx$

STEP 6

Time to integrate!
Remember the power rule?
The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$ .
Don't forget the "+ C" when doing indefinite integrals, but we don't need it for definite integrals like this one!

STEP 7

So, let's integrate our function term by term: $\int_{0}^{1} (x^2 + x + 1) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_0^1$

STEP 8

Now, we plug in our interval endpoints!
First, plug in $x = \mathbf{1}$ : $\frac{1^3}{3} + \frac{1^2}{2} + 1 = \frac{1}{3} + \frac{1}{2} + 1 = \frac{2}{6} + \frac{3}{6} + \frac{6}{6} = \frac{\mathbf{11}}{6}$ Then, plug in $x = \mathbf{0}$ (which will just give us zero in this case): $\frac{0^3}{3} + \frac{0^2}{2} + 0 = \mathbf{0}$ Subtract the second result from the first: $\frac{11}{6} - 0 = \frac{\mathbf{11}}{6}$

STEP 9

Remember that factor of $\frac{1}{1-0} = \mathbf{1}$ out front?
Let's multiply our integral result by that: $1 \cdot \frac{11}{6} = \frac{\mathbf{11}}{6}$

STEP 10

So, the average value of $f(x)$ over the interval $[0, 1]$ is $\frac{11}{6}$ .
As a decimal, that's approximately $\mathbf{1.83}$ when rounded to two decimal places.

STEP 11

The average value of $f(x) = x^2 + x + 1$ over the interval $[0, 1]$ is $\frac{11}{6}$ , which is approximately $1.83$ .

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QuestionCompute the average of $f(x)=x^{2}+x+1$ over the interval $[0,1]$ . If your answer is not a whole number, round accurate to at least two decimal places.
Moving to another question will save this response.

Studdy Solution

STEP 1

What is this asking? Find the average value of a parabola-shaped function between two points. Watch out! Don't just plug in the endpoints!
We need the average value of the function across the entire interval.

STEP 2

1. Define the function and interval
2. Set up the average value formula
3. Evaluate the integral
4. Calculate the result

STEP 3

Alright, so we've got our function $f(x) = x^2 + x + 1$ , which makes a lovely U-shaped parabola!
We're looking at it between $x = \mathbf{0}$ and $x = \mathbf{1}$ , so that's our interval!

STEP 4

STEP 5

In our case, $a = \mathbf{0}$ and $b = \mathbf{1}$ , and our function is $f(x) = x^2 + x + 1$ .
So, plugging those values into our formula, we get: $\frac{1}{1 - 0} \cdot \int_{0}^{1} (x^2 + x + 1) \, dx$

STEP 6

Time to integrate!
Remember the power rule?
The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$ .
Don't forget the "+ C" when doing indefinite integrals, but we don't need it for definite integrals like this one!

STEP 7

So, let's integrate our function term by term: $\int_{0}^{1} (x^2 + x + 1) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_0^1$

STEP 8

STEP 9

Remember that factor of $\frac{1}{1-0} = \mathbf{1}$ out front?
Let's multiply our integral result by that: $1 \cdot \frac{11}{6} = \frac{\mathbf{11}}{6}$

STEP 10

So, the average value of $f(x)$ over the interval $[0, 1]$ is $\frac{11}{6}$ .
As a decimal, that's approximately $\mathbf{1.83}$ when rounded to two decimal places.

STEP 11

The average value of $f(x) = x^2 + x + 1$ over the interval $[0, 1]$ is $\frac{11}{6}$ , which is approximately $1.83$ .

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QuestionCompute the average of f(x)=x2+x+1f(x)=x^{2}+x+1f(x)=x2+x+1 over the interval [0,1][0,1][0,1]. If your answer is not a whole number, round accurate to at least two decimal places.Moving to another question will save this response.

Studdy Solution

STEP 1

What is this asking? Find the average value of a parabola-shaped function between two points. Watch out! Don't just plug in the endpoints!We need the *average* value of the function across the entire interval.

STEP 2

1. Define the function and interval2. Set up the average value formula3. Evaluate the integral4. Calculate the result

STEP 3

Alright, so we've got our function f(x)=x2+x+1f(x) = x^2 + x + 1f(x)=x2+x+1, which makes a lovely U-shaped parabola!We're looking at it between x=0x = \mathbf{0}x=0 and x=1x = \mathbf{1}x=1, so that's our **interval**!

STEP 4

STEP 5

In our case, a=0a = \mathbf{0}a=0 and b=1b = \mathbf{1}b=1, and our function is f(x)=x2+x+1f(x) = x^2 + x + 1f(x)=x2+x+1.So, plugging those values into our formula, we get: 11−0⋅∫01(x2+x+1) dx \frac{1}{1 - 0} \cdot \int_{0}^{1} (x^2 + x + 1) \, dx 1−01​⋅∫01​(x2+x+1)dx

STEP 6

Time to integrate!Remember the power rule?The integral of xnx^nxn is xn+1n+1\frac{x^{n+1}}{n+1}n+1xn+1​.Don't forget the "+ C" when doing indefinite integrals, but we don't need it for definite integrals like this one!

STEP 7

So, let's integrate our function term by term: ∫01(x2+x+1) dx=[x33+x22+x]01 \int_{0}^{1} (x^2 + x + 1) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_0^1 ∫01​(x2+x+1)dx=[3x3​+2x2​+x]01​

STEP 8

STEP 9

Remember that factor of 11−0=1\frac{1}{1-0} = \mathbf{1}1−01​=1 out front?Let's multiply our integral result by that: 1⋅116=116 1 \cdot \frac{11}{6} = \frac{\mathbf{11}}{6} 1⋅611​=611​

STEP 10

So, the average value of f(x)f(x)f(x) over the interval [0,1][0, 1][0,1] is 116\frac{11}{6}611​.As a decimal, that's approximately 1.83\mathbf{1.83}1.83 when rounded to two decimal places.

STEP 11

The average value of f(x)=x2+x+1f(x) = x^2 + x + 1f(x)=x2+x+1 over the interval [0,1][0, 1][0,1] is 116\frac{11}{6}611​, which is approximately 1.831.831.83.

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QuestionCompute the average of $f(x)=x^{2}+x+1$ over the interval $[0,1]$ . If your answer is not a whole number, round accurate to at least two decimal places.
Moving to another question will save this response.

What is this asking? Find the average value of a parabola-shaped function between two points. Watch out! Don't just plug in the endpoints!
We need the average value of the function across the entire interval.

1. Define the function and interval
2. Set up the average value formula
3. Evaluate the integral
4. Calculate the result

Alright, so we've got our function $f(x) = x^2 + x + 1$ , which makes a lovely U-shaped parabola!
We're looking at it between $x = \mathbf{0}$ and $x = \mathbf{1}$ , so that's our interval!

In our case, $a = \mathbf{0}$ and $b = \mathbf{1}$ , and our function is $f(x) = x^2 + x + 1$ .
So, plugging those values into our formula, we get: $\frac{1}{1 - 0} \cdot \int_{0}^{1} (x^2 + x + 1) \, dx$

Time to integrate!
Remember the power rule?
The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$ .
Don't forget the "+ C" when doing indefinite integrals, but we don't need it for definite integrals like this one!

So, let's integrate our function term by term: $\int_{0}^{1} (x^2 + x + 1) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_0^1$

Remember that factor of $\frac{1}{1-0} = \mathbf{1}$ out front?
Let's multiply our integral result by that: $1 \cdot \frac{11}{6} = \frac{\mathbf{11}}{6}$

So, the average value of $f(x)$ over the interval $[0, 1]$ is $\frac{11}{6}$ .
As a decimal, that's approximately $\mathbf{1.83}$ when rounded to two decimal places.

The average value of $f(x) = x^2 + x + 1$ over the interval $[0, 1]$ is $\frac{11}{6}$ , which is approximately $1.83$ .