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### 3.1 Evaluate the definite integral:
[tex]$ \int_0^1 \frac{x^2}{x^3+1}\, dx $[/tex]
Given integral can be evaluated using methods of integration, and we find:
[tex]$ 0.231049060186648 $[/tex]
### 3.2 Given:
[tex]$ y = x^2 \quad \text{and} \quad y = 2x $[/tex]
#### 3.2.1 Calculate the coordinates of the points of intersection.
To find the points where the curves intersect each other, we set the equations equal to each other:
[tex]$ x^2 = 2x $[/tex]
Re-arranging, we get:
[tex]$ x^2 - 2x = 0 $[/tex]
Factoring the equation:
[tex]$ x(x - 2) = 0 $[/tex]
This gives us:
[tex]$ x = 0 \quad \text{or} \quad x = 2 $[/tex]
For [tex]\( x = 0 \)[/tex]:
[tex]$ y = 0^2 = 0 $[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]$ y = 2^2 = 4 $[/tex]
Therefore, the points of intersection are:
[tex]$ (0.0, 0.0) \quad \text{and} \quad (2.0, 4.0) $[/tex]
#### 3.2.2 Make a neat sketch to show the enclosed area, the representative strip and the point of intersection.
Comment Only: Use a graph plotting tool to draw the curves [tex]\( y = x^2 \)[/tex] and [tex]\( y = 2x \)[/tex]. Mark the points of intersection (0,0) and (2,4). Shade the area enclosed between these curves and draw a representative vertical strip in this region.
#### 3.2.3 Calculate the magnitude of the area in QUESTION 3.2.2
The area enclosed between the curves can be found by integrating the difference of the functions from [tex]\( x = 0 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]$ \int_0^2 (2x - x^2) \, dx $[/tex]
Evaluating this integral, we get:
[tex]$ 1.33333333333333 \quad or \quad \frac{4}{3} $[/tex]
#### 3.2.4 Calculate the volume of the solid of revolution formed when the area in QUESTION 3.2.2 is rotated about the [tex]\( x \)[/tex]-axis.
To find the volume of the solid of revolution when the area is rotated about the [tex]\( x \)[/tex]-axis, we use the method of disks or washers. The formula for the volume is:
[tex]$ V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx $[/tex]
Where [tex]\( R(x) \)[/tex] is the outer radius and [tex]\( r(x) \)[/tex] is the inner radius. Here:
[tex]$ R(x) = 2x \quad \text{and} \quad r(x) = x^2 $[/tex]
Thus, we need to evaluate:
[tex]$ V = \pi \int_0^2 ((2x)^2 - (x^2)^2) \, dx $[/tex]
Evaluating this integral, we get:
[tex]$ 13.4041286553165 $[/tex]
### 3.3 Calculate the second moment of area of a rectangular lamina with sides [tex]\( 8 \)[/tex] cm [tex]\( \times \)[/tex] [tex]\( 4 \)[/tex] cm about a [tex]\( 4 \)[/tex] cm side.
The formula for the second moment of area (also known as the area moment of inertia) about the base of a rectangle of width [tex]\( b \)[/tex] and height [tex]\( h \)[/tex] is given by:
[tex]$ I = \frac{b h^3}{12} $[/tex]
Given [tex]\( b = 4 \)[/tex] cm and [tex]\( h = 8 \)[/tex] cm, we have:
[tex]$ I = \frac{4 \cdot 8^3}{12} $[/tex]
Calculating this, we get:
[tex]$ 170.66666666666666 \quad \text{cm}^4 $[/tex]
### 3.1 Evaluate the definite integral:
[tex]$ \int_0^1 \frac{x^2}{x^3+1}\, dx $[/tex]
Given integral can be evaluated using methods of integration, and we find:
[tex]$ 0.231049060186648 $[/tex]
### 3.2 Given:
[tex]$ y = x^2 \quad \text{and} \quad y = 2x $[/tex]
#### 3.2.1 Calculate the coordinates of the points of intersection.
To find the points where the curves intersect each other, we set the equations equal to each other:
[tex]$ x^2 = 2x $[/tex]
Re-arranging, we get:
[tex]$ x^2 - 2x = 0 $[/tex]
Factoring the equation:
[tex]$ x(x - 2) = 0 $[/tex]
This gives us:
[tex]$ x = 0 \quad \text{or} \quad x = 2 $[/tex]
For [tex]\( x = 0 \)[/tex]:
[tex]$ y = 0^2 = 0 $[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]$ y = 2^2 = 4 $[/tex]
Therefore, the points of intersection are:
[tex]$ (0.0, 0.0) \quad \text{and} \quad (2.0, 4.0) $[/tex]
#### 3.2.2 Make a neat sketch to show the enclosed area, the representative strip and the point of intersection.
Comment Only: Use a graph plotting tool to draw the curves [tex]\( y = x^2 \)[/tex] and [tex]\( y = 2x \)[/tex]. Mark the points of intersection (0,0) and (2,4). Shade the area enclosed between these curves and draw a representative vertical strip in this region.
#### 3.2.3 Calculate the magnitude of the area in QUESTION 3.2.2
The area enclosed between the curves can be found by integrating the difference of the functions from [tex]\( x = 0 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]$ \int_0^2 (2x - x^2) \, dx $[/tex]
Evaluating this integral, we get:
[tex]$ 1.33333333333333 \quad or \quad \frac{4}{3} $[/tex]
#### 3.2.4 Calculate the volume of the solid of revolution formed when the area in QUESTION 3.2.2 is rotated about the [tex]\( x \)[/tex]-axis.
To find the volume of the solid of revolution when the area is rotated about the [tex]\( x \)[/tex]-axis, we use the method of disks or washers. The formula for the volume is:
[tex]$ V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx $[/tex]
Where [tex]\( R(x) \)[/tex] is the outer radius and [tex]\( r(x) \)[/tex] is the inner radius. Here:
[tex]$ R(x) = 2x \quad \text{and} \quad r(x) = x^2 $[/tex]
Thus, we need to evaluate:
[tex]$ V = \pi \int_0^2 ((2x)^2 - (x^2)^2) \, dx $[/tex]
Evaluating this integral, we get:
[tex]$ 13.4041286553165 $[/tex]
### 3.3 Calculate the second moment of area of a rectangular lamina with sides [tex]\( 8 \)[/tex] cm [tex]\( \times \)[/tex] [tex]\( 4 \)[/tex] cm about a [tex]\( 4 \)[/tex] cm side.
The formula for the second moment of area (also known as the area moment of inertia) about the base of a rectangle of width [tex]\( b \)[/tex] and height [tex]\( h \)[/tex] is given by:
[tex]$ I = \frac{b h^3}{12} $[/tex]
Given [tex]\( b = 4 \)[/tex] cm and [tex]\( h = 8 \)[/tex] cm, we have:
[tex]$ I = \frac{4 \cdot 8^3}{12} $[/tex]
Calculating this, we get:
[tex]$ 170.66666666666666 \quad \text{cm}^4 $[/tex]
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