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Find the area of $\newline$ the curved trapezoid bounded by the lines $y=3x+6$ , $y=0$ , $x=-1$ , $x=2$ .

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Find the magnitude of a three-dimensional vector

Full solution

Q. Find the area of

\newline

the curved trapezoid bounded by the lines

y=3x+6

,

y=0

,

x=-1

,

x=2

.

Find Intersection Points: First, we need to find the points of intersection between the lines to determine the vertices of the trapezoid. $\newline$ For $y=0$ and $x=-1$ , the point is $(-1,0)$ . $\newline$ For $y=0$ and $x=2$ , the point is $(2,0)$ . $\newline$ For $y=3x+6$ and $x=-1$ , $y=3(-1)+6=3$ , so the point is $(-1,3)$ . $\newline$ For $y=3x+6$ and $x=2$ , $x=-1$ $2$ , so the point is $x=-1$ $3$ .
Calculate Trapezoid Area: Now, we calculate the area of the trapezoid. The formula for the area of a trapezoid is $A = \frac{1}{2} \times (b_1 + b_2) \times h$ , where $b_1$ and $b_2$ are the bases and $h$ is the height. Here, $b_1$ is the length from $(-1,0)$ to $(2,0)$ , which is $3$ units. $b_2$ is the length from $(-1,3)$ to $b_1$ $0$ , which we find by calculating the distance between these two points.
Find Distance Between Points: To find the distance between $(-1,3)$ and $(2,12)$ , we use the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ . So, $d = \sqrt{(2-(-1))^2 + (12-3)^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90}$ .
Calculate Trapezoid Height: The height $h$ of the trapezoid is the distance between $y=0$ and $y=3x+6$ along the $x=-1$ or $x=2$ line. $\newline$ Since these lines are vertical, the height is simply the difference in $y$ -values at $x=-1$ or $x=2$ . $\newline$ For $x=-1$ , the $y$ -values are $y=0$ $0$ and $y=0$ $1$ , so $y=0$ $2$ . $\newline$ For $x=2$ , the $y$ -values are $y=0$ $0$ and $y=0$ $6$ , so $y=0$ $7$ . $\newline$ We made a mistake here; the height should be consistent, so we need to choose one $y=0$ $8$ -value to calculate the height.

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