Area bounded by \(y=|x-5|\) and x-axis between \(x = 2\) and \(x = 4\) is:
To find the area bounded by the curve \(y=|x-5|\) and the x-axis between \(x=2\) and \(x=4\), we first need to understand the behavior of the absolute value function \(y=|x-5|\).
This function can be rewritten in piecewise form as:
\(y=x-5\) when \(x\geq5\) and \(y=-(x-5)=5-x\) when \(x<5\).
Since \(2\leq x\leq4\), in this interval \(x<5\), so we use \(y=5-x\).
We can now find the area under the curve \(y=5-x\) from \(x=2\) to \(x=4\) by calculating the definite integral:
\(\text{Area} = \int_{2}^{4}(5-x) \, dx\)
Calculate the integral: \(\int(5-x) \, dx = 5x - \frac{x^2}{2} + C\).
Evaluate it from 2 to 4:
\(\text{Area} = \left[5x - \frac{x^2}{2}\right]_{2}^{4}\)
\(\text{Area} = (5(4) - \frac{4^2}{2}) - (5(2) - \frac{2^2}{2})\)
\(\text{Area} = (20 - 8) - (10 - 2)\)
\(\text{Area} = 12 - 8\)
\(\text{Area} = 4\)
The area bounded by \(y=|x-5|\) and the x-axis from \(x=2\) to \(x=4\) is \(\boxed{4}\).
The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \)} and \( x = 2 \) is:
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to: