Math / Calculus

QuestionUse geometry to evaluate the definite integral. $\int_{0}^{2} 2 d x$

Studdy Solution

STEP 1

What is this asking? We need to find the area under the line $y = 2$ from $x = 0$ to $x = 2$ . Watch out! Don't forget that a definite integral represents the signed area.
Areas below the x-axis are negative!
Luckily, our function is always positive, so we don't have to worry about that here.

STEP 2

1. Define the problem geometrically
2. Calculate the area

STEP 3

Imagine the line $y = 2$ .
It's a nice, horizontal line, cruising two units above the x-axis.
Now, we're looking at the area underneath this line, but above the x-axis, between $x = 0$ and $x = 2$ .
What shape does that make?

STEP 4

It's a rectangle!
The base of the rectangle runs along the x-axis from $x = 0$ to $x = 2$ , so its length is $2 - 0 = 2$ .
The height of the rectangle is the constant value of our function, which is $2$ .

STEP 5

The area of a rectangle is simply base times height.

STEP 6

In our case, the base is $2$ and the height is $2$ .
So, the area is: $\text{Area} = \text{base} \cdot \text{height} = 2 \cdot 2 = 4$

STEP 7

The definite integral $\int_{0}^{2} 2 \, dx$ is equal to 4.

Was this helpful?

QuestionUse geometry to evaluate the definite integral. $\int_{0}^{2} 2 d x$

Studdy Solution

STEP 1

What is this asking? We need to find the area under the line $y = 2$ from $x = 0$ to $x = 2$ . Watch out! Don't forget that a definite integral represents the signed area.
Areas below the x-axis are negative!
Luckily, our function is always positive, so we don't have to worry about that here.

STEP 2

1. Define the problem geometrically
2. Calculate the area

STEP 3

Imagine the line $y = 2$ .
It's a nice, horizontal line, cruising two units above the x-axis.
Now, we're looking at the area underneath this line, but above the x-axis, between $x = 0$ and $x = 2$ .
What shape does that make?

STEP 4

It's a rectangle!
The base of the rectangle runs along the x-axis from $x = 0$ to $x = 2$ , so its length is $2 - 0 = 2$ .
The height of the rectangle is the constant value of our function, which is $2$ .

STEP 5

The area of a rectangle is simply base times height.

STEP 6

In our case, the base is $2$ and the height is $2$ .
So, the area is: $\text{Area} = \text{base} \cdot \text{height} = 2 \cdot 2 = 4$

STEP 7

The definite integral $\int_{0}^{2} 2 \, dx$ is equal to 4.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionUse geometry to evaluate the definite integral. ∫022dx\int_{0}^{2} 2 d x∫02​2dx

Studdy Solution

STEP 1

STEP 2

1. Define the problem geometrically2. Calculate the area

STEP 3

Imagine the line y=2y = 2y=2.It's a nice, horizontal line, cruising two units above the x-axis.Now, we're looking at the area underneath this line, but above the x-axis, between x=0x = 0x=0 and x=2x = 2x=2.What shape does that make?

STEP 4

It's a **rectangle**!The **base** of the rectangle runs along the x-axis from x=0x = 0x=0 to x=2x = 2x=2, so its length is 2−0=22 - 0 = 22−0=2.The **height** of the rectangle is the constant value of our function, which is 222.

STEP 5

The area of a rectangle is simply **base** times **height**.

STEP 6

In our case, the **base** is 222 and the **height** is 222.So, the area is: Area=base⋅height=2⋅2=4\text{Area} = \text{base} \cdot \text{height} = 2 \cdot 2 = 4Area=base⋅height=2⋅2=4

STEP 7

The definite integral ∫022 dx\int_{0}^{2} 2 \, dx∫02​2dx is equal to **4**.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionUse geometry to evaluate the definite integral. $\int_{0}^{2} 2 d x$

1. Define the problem geometrically
2. Calculate the area

Imagine the line $y = 2$ .
It's a nice, horizontal line, cruising two units above the x-axis.
Now, we're looking at the area underneath this line, but above the x-axis, between $x = 0$ and $x = 2$ .
What shape does that make?

It's a rectangle!
The base of the rectangle runs along the x-axis from $x = 0$ to $x = 2$ , so its length is $2 - 0 = 2$ .
The height of the rectangle is the constant value of our function, which is $2$ .

The area of a rectangle is simply base times height.

In our case, the base is $2$ and the height is $2$ .
So, the area is: $\text{Area} = \text{base} \cdot \text{height} = 2 \cdot 2 = 4$

The definite integral $\int_{0}^{2} 2 \, dx$ is equal to 4.