We need to find the area enclosed by the function $y = 8 - x$ and the coordinate axes (x-axis and y-axis) specifically within the first quadrant ($x \ge 0, y \ge 0$).
The boundaries are:
To find the area, we first determine where the line intersects the coordinate axes:
The region bounded by the line and the axes in the first quadrant is a right-angled triangle with vertices at the origin $(0, 0)$, the x-intercept $(8, 0)$, and the y-intercept $(0, 8)$.
The base of the triangle lies along the x-axis and has a length of 8 units. The height of the triangle lies along the y-axis and also has a length of 8 units.
The formula for the area of a triangle is:
$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $
Substitute the base and height values:
$ \text{Area} = \frac{1}{2} \times 8 \times 8 $
$ \text{Area} = \frac{1}{2} \times 64 $
$ \text{Area} = 32 $
The area bounded by the function $y = 8 - x$ and the coordinate axes in the first quadrant is 32 square units.
The equation of a closed curve in two-dimensional polar coordinates is given by $r = \frac{2}{\sqrt{\pi}}(1 - \sin \theta)$. The area enclosed by the curve is ______ (answer in integer).
The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 decimal places)
Two straight lines pass through the origin $(x_0, y_0) = (0,0)$. One of them passes through the point $(x_1, y_1) = (1,3)$ and the other passes through the point $(x_2, y_2) = (1,2)$.
What is the area enclosed between the straight lines in the interval $[0, 1]$ on the x-axis?
Consider the equation for a curve, $y = f(x) = x^2 + x$.
The area enclosed by the curve, the x -axis ($y = 0$ line); the vertical lines passing through $x = 1$ and $x = 2$ is _________ (rounded off to 2 decimal places)