Math / Calculus

QuestionFor the region formed by the functions $f(x)=x^{2}-2 x-2$ and $g(x)=5$ on the interval $[-1,2]$ , use definite integrals to find the area of the region. Answer: The area is $\square$ Hint: Follow Example 2.

Studdy Solution

STEP 1

What is this asking? We need to find the area trapped between two curves, $f(x)$ and $g(x)$ , within a specific range of $x$ values. Watch out! Make sure you figure out which function is on top before setting up the integral!
Getting the order wrong will give you a negative area, which isn't physically possible.

STEP 2

1. Visualize the Problem
2. Determine the Top Function
3. Set Up the Integral
4. Evaluate the Integral

STEP 3

Let's imagine what's going on!
We've got a parabola $f(x) = x^2 - 2x - 2$ and a horizontal line $g(x) = 5$ .
We're looking at the area between them from $x = -1$ to $x = 2$ .

STEP 4

Which function has larger $y$ values in the given interval?
Let's test a point!
Let's pick $x = 0$ , which is nicely in the middle of our interval.

STEP 5

We get $f(0) = 0^2 - 2 \cdot 0 - 2 = -2$ and $g(0) = 5$ .
Since $5 > -2$ , it looks like $g(x)$ is above $f(x)$ in our interval.

STEP 6

The area between two curves is found by integrating the difference of the top function and the bottom function.
Since $g(x)$ is on top, we'll integrate $(g(x) - f(x))$ from $x = -1$ to $x = 2$ .

STEP 7

Our integral looks like this: $\int_{-1}^{2} (g(x) - f(x)) \, dx = \int_{-1}^{2} (5 - (x^2 - 2x - 2)) \, dx$

STEP 8

Let's simplify the integrand: $\int_{-1}^{2} (5 - x^2 + 2x + 2) \, dx = \int_{-1}^{2} (-x^2 + 2x + 7) \, dx$

STEP 9

Power Up! Apply the power rule for integration: $\int_{-1}^{2} (-x^2 + 2x + 7) \, dx = \left[ -\frac{1}{3}x^3 + x^2 + 7x \right]_{-1}^{2}$

STEP 10

Plug and Chug! Substitute the upper limit ( $x = 2$ ) and the lower limit ( $x = -1$ ): $\left( -\frac{1}{3}(2)^3 + (2)^2 + 7(2) \right) - \left( -\frac{1}{3}(-1)^3 + (-1)^2 + 7(-1) \right)$

STEP 11

Simplify! $\left( -\frac{8}{3} + 4 + 14 \right) - \left( \frac{1}{3} + 1 - 7 \right) = \left( \frac{-8 + 12 + 42}{3} \right) - \left( \frac{1 + 3 - 21}{3} \right)$

STEP 12

$= \frac{46}{3} - \frac{-17}{3} = \frac{46 + 17}{3} = \frac{63}{3} = 21$

STEP 13

The area of the region between the curves is 21 square units.

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QuestionFor the region formed by the functions $f(x)=x^{2}-2 x-2$ and $g(x)=5$ on the interval $[-1,2]$ , use definite integrals to find the area of the region. Answer: The area is $\square$ Hint: Follow Example 2.

Studdy Solution

STEP 1

STEP 2

1. Visualize the Problem
2. Determine the Top Function
3. Set Up the Integral
4. Evaluate the Integral

STEP 3

Let's imagine what's going on!
We've got a parabola $f(x) = x^2 - 2x - 2$ and a horizontal line $g(x) = 5$ .
We're looking at the area between them from $x = -1$ to $x = 2$ .

STEP 4

Which function has larger $y$ values in the given interval?
Let's test a point!
Let's pick $x = 0$ , which is nicely in the middle of our interval.

STEP 5

We get $f(0) = 0^2 - 2 \cdot 0 - 2 = -2$ and $g(0) = 5$ .
Since $5 > -2$ , it looks like $g(x)$ is above $f(x)$ in our interval.

STEP 6

The area between two curves is found by integrating the difference of the top function and the bottom function.
Since $g(x)$ is on top, we'll integrate $(g(x) - f(x))$ from $x = -1$ to $x = 2$ .

STEP 7

Our integral looks like this: $\int_{-1}^{2} (g(x) - f(x)) \, dx = \int_{-1}^{2} (5 - (x^2 - 2x - 2)) \, dx$

STEP 8

Let's simplify the integrand: $\int_{-1}^{2} (5 - x^2 + 2x + 2) \, dx = \int_{-1}^{2} (-x^2 + 2x + 7) \, dx$

STEP 9

Power Up! Apply the power rule for integration: $\int_{-1}^{2} (-x^2 + 2x + 7) \, dx = \left[ -\frac{1}{3}x^3 + x^2 + 7x \right]_{-1}^{2}$

STEP 10

Plug and Chug! Substitute the upper limit ( $x = 2$ ) and the lower limit ( $x = -1$ ): $\left( -\frac{1}{3}(2)^3 + (2)^2 + 7(2) \right) - \left( -\frac{1}{3}(-1)^3 + (-1)^2 + 7(-1) \right)$

STEP 11

Simplify! $\left( -\frac{8}{3} + 4 + 14 \right) - \left( \frac{1}{3} + 1 - 7 \right) = \left( \frac{-8 + 12 + 42}{3} \right) - \left( \frac{1 + 3 - 21}{3} \right)$

STEP 12

$= \frac{46}{3} - \frac{-17}{3} = \frac{46 + 17}{3} = \frac{63}{3} = 21$

STEP 13

The area of the region between the curves is 21 square units.

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QuestionFor the region formed by the functions f(x)=x2−2x−2f(x)=x^{2}-2 x-2f(x)=x2−2x−2 and g(x)=5g(x)=5g(x)=5 on the interval [−1,2][-1,2][−1,2], use definite integrals to find the area of the region. Answer: The area is □\square□ Hint: Follow Example 2.

Studdy Solution

STEP 1

STEP 2

1. Visualize the Problem2. Determine the Top Function3. Set Up the Integral4. Evaluate the Integral

STEP 3

Let's imagine what's going on!We've got a parabola f(x)=x2−2x−2f(x) = x^2 - 2x - 2f(x)=x2−2x−2 and a horizontal line g(x)=5g(x) = 5g(x)=5.We're looking at the area between them from x=−1x = -1x=−1 to x=2x = 2x=2.

STEP 4

Which function has larger yyy values in the given interval?Let's test a point!Let's pick x=0x = 0x=0, which is nicely in the middle of our interval.

STEP 5

We get f(0)=02−2⋅0−2=−2f(0) = 0^2 - 2 \cdot 0 - 2 = -2f(0)=02−2⋅0−2=−2 and g(0)=5g(0) = 5g(0)=5.Since 5>−25 > -25>−2, it looks like g(x)g(x)g(x) is above f(x)f(x)f(x) in our interval.

STEP 6

The area between two curves is found by integrating the *difference* of the top function and the bottom function.Since g(x)g(x)g(x) is on top, we'll integrate (g(x)−f(x))(g(x) - f(x))(g(x)−f(x)) from x=−1x = -1x=−1 to x=2x = 2x=2.

STEP 7

Our integral looks like this: ∫−12(g(x)−f(x)) dx=∫−12(5−(x2−2x−2)) dx \int_{-1}^{2} (g(x) - f(x)) \, dx = \int_{-1}^{2} (5 - (x^2 - 2x - 2)) \, dx ∫−12​(g(x)−f(x))dx=∫−12​(5−(x2−2x−2))dx

STEP 8

Let's simplify the integrand: ∫−12(5−x2+2x+2) dx=∫−12(−x2+2x+7) dx \int_{-1}^{2} (5 - x^2 + 2x + 2) \, dx = \int_{-1}^{2} (-x^2 + 2x + 7) \, dx ∫−12​(5−x2+2x+2)dx=∫−12​(−x2+2x+7)dx

STEP 9

**Power Up!** Apply the power rule for integration: ∫−12(−x2+2x+7) dx=[−13x3+x2+7x]−12 \int_{-1}^{2} (-x^2 + 2x + 7) \, dx = \left[ -\frac{1}{3}x^3 + x^2 + 7x \right]_{-1}^{2} ∫−12​(−x2+2x+7)dx=[−31​x3+x2+7x]−12​

STEP 10

STEP 11

**Simplify!** (−83+4+14)−(13+1−7)=(−8+12+423)−(1+3−213) \left( -\frac{8}{3} + 4 + 14 \right) - \left( \frac{1}{3} + 1 - 7 \right) = \left( \frac{-8 + 12 + 42}{3} \right) - \left( \frac{1 + 3 - 21}{3} \right) (−38​+4+14)−(31​+1−7)=(3−8+12+42​)−(31+3−21​)

STEP 12

=463−−173=46+173=633=21 = \frac{46}{3} - \frac{-17}{3} = \frac{46 + 17}{3} = \frac{63}{3} = 21 =346​−3−17​=346+17​=363​=21

STEP 13

The area of the region between the curves is **21 square units**.

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QuestionFor the region formed by the functions $f(x)=x^{2}-2 x-2$ and $g(x)=5$ on the interval $[-1,2]$ , use definite integrals to find the area of the region. Answer: The area is $\square$ Hint: Follow Example 2.

1. Visualize the Problem
2. Determine the Top Function
3. Set Up the Integral
4. Evaluate the Integral

Let's imagine what's going on!
We've got a parabola $f(x) = x^2 - 2x - 2$ and a horizontal line $g(x) = 5$ .
We're looking at the area between them from $x = -1$ to $x = 2$ .

Which function has larger $y$ values in the given interval?
Let's test a point!
Let's pick $x = 0$ , which is nicely in the middle of our interval.

We get $f(0) = 0^2 - 2 \cdot 0 - 2 = -2$ and $g(0) = 5$ .
Since $5 > -2$ , it looks like $g(x)$ is above $f(x)$ in our interval.

The area between two curves is found by integrating the difference of the top function and the bottom function.
Since $g(x)$ is on top, we'll integrate $(g(x) - f(x))$ from $x = -1$ to $x = 2$ .

Our integral looks like this: $\int_{-1}^{2} (g(x) - f(x)) \, dx = \int_{-1}^{2} (5 - (x^2 - 2x - 2)) \, dx$

Let's simplify the integrand: $\int_{-1}^{2} (5 - x^2 + 2x + 2) \, dx = \int_{-1}^{2} (-x^2 + 2x + 7) \, dx$

Power Up! Apply the power rule for integration: $\int_{-1}^{2} (-x^2 + 2x + 7) \, dx = \left[ -\frac{1}{3}x^3 + x^2 + 7x \right]_{-1}^{2}$

Simplify! $\left( -\frac{8}{3} + 4 + 14 \right) - \left( \frac{1}{3} + 1 - 7 \right) = \left( \frac{-8 + 12 + 42}{3} \right) - \left( \frac{1 + 3 - 21}{3} \right)$

$= \frac{46}{3} - \frac{-17}{3} = \frac{46 + 17}{3} = \frac{63}{3} = 21$

The area of the region between the curves is 21 square units.