Math / Calculus

Question34. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) $\int_{0}^{2} g(x) d x$ (b) $\int_{2}^{6} g(x) d x$ (c) $\int_{0}^{7} g(x) d x$

Studdy Solution

STEP 1

What is this asking? We need to find the area under the graph of $g(x)$ for three different intervals. Watch out! Areas below the x-axis are considered negative!
Also, remember the formula for the area of a circle is $\pi \cdot r^2$ .

STEP 2

1. Area of the triangle
2. Area of the semicircle
3. Area of the second triangle
4. Sum it all up!

STEP 3

Let's start with the interval $[0, 2]$ .
This part of the graph is a triangle!
We know the area of a triangle is $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .

STEP 4

The base of our triangle goes from $x = 0$ to $x = 2$ , so it has a length of $2 - 0 = 2$ .
The height is $4$ .

STEP 5

So, the area of this triangle is $\frac{1}{2} \cdot 2 \cdot 4 = 4.$

STEP 6

Next up is the interval $[2, 6]$ .
This part of the graph is a semicircle below the x-axis.
Remember, the area of a full circle is $\pi \cdot r^2$ .
A semicircle is half of that, so its area is $\frac{1}{2} \cdot \pi \cdot r^2$ .

STEP 7

Looking at the graph, the radius of the semicircle is $2$ .

STEP 8

So, the area of the semicircle is $\frac{1}{2} \cdot \pi \cdot 2^2 = \frac{1}{2} \cdot \pi \cdot 4 = 2\pi.$ Since it's below the x-axis, this area is considered negative, so we have $-2\pi$ .

STEP 9

Finally, we have the interval $[6, 7]$ .
This is another triangle!

STEP 10

The base of this triangle goes from $x = 6$ to $x = 7$ , so it has a length of $7 - 6 = 1$ .
The height is $2$ .

STEP 11

The area of this triangle is $\frac{1}{2} \cdot 1 \cdot 2 = 1.$

STEP 12

For the integral from $0$ to $7$ , we add the areas we found: $4$ (the first triangle), $-2\pi$ (the semicircle), and $1$ (the second triangle).

STEP 13

$\int_{0}^{7} g(x) \, dx = 4 + (-2\pi) + 1 = 5 - 2\pi.$

STEP 14

(a) $\int_{0}^{2} g(x) \, dx = 4$ (b) $\int_{2}^{6} g(x) \, dx = -2\pi$ (c) $\int_{0}^{7} g(x) \, dx = 5 - 2\pi$

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Question34. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) $\int_{0}^{2} g(x) d x$ (b) $\int_{2}^{6} g(x) d x$ (c) $\int_{0}^{7} g(x) d x$

Studdy Solution

STEP 1

What is this asking? We need to find the area under the graph of $g(x)$ for three different intervals. Watch out! Areas below the x-axis are considered negative!
Also, remember the formula for the area of a circle is $\pi \cdot r^2$ .

STEP 2

1. Area of the triangle
2. Area of the semicircle
3. Area of the second triangle
4. Sum it all up!

STEP 3

Let's start with the interval $[0, 2]$ .
This part of the graph is a triangle!
We know the area of a triangle is $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .

STEP 4

The base of our triangle goes from $x = 0$ to $x = 2$ , so it has a length of $2 - 0 = 2$ .
The height is $4$ .

STEP 5

So, the area of this triangle is $\frac{1}{2} \cdot 2 \cdot 4 = 4.$

STEP 6

Next up is the interval $[2, 6]$ .
This part of the graph is a semicircle below the x-axis.
Remember, the area of a full circle is $\pi \cdot r^2$ .
A semicircle is half of that, so its area is $\frac{1}{2} \cdot \pi \cdot r^2$ .

STEP 7

Looking at the graph, the radius of the semicircle is $2$ .

STEP 8

So, the area of the semicircle is $\frac{1}{2} \cdot \pi \cdot 2^2 = \frac{1}{2} \cdot \pi \cdot 4 = 2\pi.$ Since it's below the x-axis, this area is considered negative, so we have $-2\pi$ .

STEP 9

Finally, we have the interval $[6, 7]$ .
This is another triangle!

STEP 10

The base of this triangle goes from $x = 6$ to $x = 7$ , so it has a length of $7 - 6 = 1$ .
The height is $2$ .

STEP 11

The area of this triangle is $\frac{1}{2} \cdot 1 \cdot 2 = 1.$

STEP 12

For the integral from $0$ to $7$ , we add the areas we found: $4$ (the first triangle), $-2\pi$ (the semicircle), and $1$ (the second triangle).

STEP 13

$\int_{0}^{7} g(x) \, dx = 4 + (-2\pi) + 1 = 5 - 2\pi.$

STEP 14

(a) $\int_{0}^{2} g(x) \, dx = 4$ (b) $\int_{2}^{6} g(x) \, dx = -2\pi$ (c) $\int_{0}^{7} g(x) \, dx = 5 - 2\pi$

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Question34. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) ∫02g(x)dx\int_{0}^{2} g(x) d x∫02​g(x)dx (b) ∫26g(x)dx\int_{2}^{6} g(x) d x∫26​g(x)dx (c) ∫07g(x)dx\int_{0}^{7} g(x) d x∫07​g(x)dx

Studdy Solution

STEP 1

What is this asking? We need to find the area under the graph of g(x)g(x)g(x) for three different intervals. Watch out! Areas below the x-axis are considered negative!Also, remember the formula for the area of a circle is π⋅r2\pi \cdot r^2π⋅r2.

STEP 2

1. Area of the triangle2. Area of the semicircle3. Area of the second triangle4. Sum it all up!

STEP 3

Let's **start** with the interval [0,2][0, 2][0,2].This part of the graph is a **triangle**!We know the area of a triangle is 12⋅base⋅height \frac{1}{2} \cdot \text{base} \cdot \text{height} 21​⋅base⋅height.

STEP 4

The **base** of our triangle goes from x=0x = 0x=0 to x=2x = 2x=2, so it has a length of 2−0=22 - 0 = 22−0=2.The **height** is 444.

STEP 5

So, the area of this triangle is 12⋅2⋅4=4. \frac{1}{2} \cdot 2 \cdot 4 = 4. 21​⋅2⋅4=4.

STEP 6

Next up is the interval [2,6][2, 6][2,6].This part of the graph is a **semicircle** below the x-axis.Remember, the area of a full circle is π⋅r2\pi \cdot r^2π⋅r2.A semicircle is half of that, so its area is 12⋅π⋅r2\frac{1}{2} \cdot \pi \cdot r^221​⋅π⋅r2.

STEP 7

Looking at the graph, the **radius** of the semicircle is 222.

STEP 8

So, the area of the semicircle is 12⋅π⋅22=12⋅π⋅4=2π. \frac{1}{2} \cdot \pi \cdot 2^2 = \frac{1}{2} \cdot \pi \cdot 4 = 2\pi. 21​⋅π⋅22=21​⋅π⋅4=2π. Since it's *below* the x-axis, this area is considered **negative**, so we have −2π-2\pi−2π.

STEP 9

Finally, we have the interval [6,7][6, 7][6,7].This is another **triangle**!

STEP 10

The **base** of this triangle goes from x=6x = 6x=6 to x=7x = 7x=7, so it has a length of 7−6=17 - 6 = 17−6=1.The **height** is 222.

STEP 11

The area of this triangle is 12⋅1⋅2=1. \frac{1}{2} \cdot 1 \cdot 2 = 1. 21​⋅1⋅2=1.

STEP 12

For the integral from 000 to 777, we **add** the areas we found: 444 (the first triangle), −2π-2\pi−2π (the semicircle), and 111 (the second triangle).

STEP 13

∫07g(x) dx=4+(−2π)+1=5−2π. \int_{0}^{7} g(x) \, dx = 4 + (-2\pi) + 1 = 5 - 2\pi. ∫07​g(x)dx=4+(−2π)+1=5−2π.

STEP 14

(a) ∫02g(x) dx=4\int_{0}^{2} g(x) \, dx = 4∫02​g(x)dx=4 (b) ∫26g(x) dx=−2π\int_{2}^{6} g(x) \, dx = -2\pi∫26​g(x)dx=−2π (c) ∫07g(x) dx=5−2π\int_{0}^{7} g(x) \, dx = 5 - 2\pi∫07​g(x)dx=5−2π

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Question34. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) $\int_{0}^{2} g(x) d x$ (b) $\int_{2}^{6} g(x) d x$ (c) $\int_{0}^{7} g(x) d x$

What is this asking? We need to find the area under the graph of $g(x)$ for three different intervals. Watch out! Areas below the x-axis are considered negative!
Also, remember the formula for the area of a circle is $\pi \cdot r^2$ .

1. Area of the triangle
2. Area of the semicircle
3. Area of the second triangle
4. Sum it all up!

Let's start with the interval $[0, 2]$ .
This part of the graph is a triangle!
We know the area of a triangle is $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .

The base of our triangle goes from $x = 0$ to $x = 2$ , so it has a length of $2 - 0 = 2$ .
The height is $4$ .

So, the area of this triangle is $\frac{1}{2} \cdot 2 \cdot 4 = 4.$

Next up is the interval $[2, 6]$ .
This part of the graph is a semicircle below the x-axis.
Remember, the area of a full circle is $\pi \cdot r^2$ .
A semicircle is half of that, so its area is $\frac{1}{2} \cdot \pi \cdot r^2$ .

Looking at the graph, the radius of the semicircle is $2$ .

So, the area of the semicircle is $\frac{1}{2} \cdot \pi \cdot 2^2 = \frac{1}{2} \cdot \pi \cdot 4 = 2\pi.$ Since it's below the x-axis, this area is considered negative, so we have $-2\pi$ .

Finally, we have the interval $[6, 7]$ .
This is another triangle!

The base of this triangle goes from $x = 6$ to $x = 7$ , so it has a length of $7 - 6 = 1$ .
The height is $2$ .

The area of this triangle is $\frac{1}{2} \cdot 1 \cdot 2 = 1.$

For the integral from $0$ to $7$ , we add the areas we found: $4$ (the first triangle), $-2\pi$ (the semicircle), and $1$ (the second triangle).

$\int_{0}^{7} g(x) \, dx = 4 + (-2\pi) + 1 = 5 - 2\pi.$

(a) $\int_{0}^{2} g(x) \, dx = 4$ (b) $\int_{2}^{6} g(x) \, dx = -2\pi$ (c) $\int_{0}^{7} g(x) \, dx = 5 - 2\pi$