Question

Sketch the region y=sqrtx, y=0, x=4

Answer 1

To sketch the region defined by the equations y = sqrt(x), y = 0, and x = 4, plot the points where these equations intersect to form a right-angled triangle.

Step-by-step explanation:

To sketch the region defined by the equations y = sqrt(x), y = 0, and x = 4, you need to plot the points where these equations intersect.

The equation y = sqrt(x) represents a half-parabola that starts at the origin (0, 0) and curves upwards as x increases. The equation y = 0 represents the x-axis, a straight line with y = 0, and the equation x = 4 represents a vertical line at x = 4.

To sketch the region, mark the points where these equations intersect on a graph. You will have a right-angled triangle with one side along the x-axis starting from the origin extending to x = 4 and another side along the curve of y = sqrt(x).

Answer 2

see attached for a sketch

area = 5 1/3 square units

Step-by-step explanation:

You want the area under the square root curve, above y=0, from x=0 to x=4.

Area

The area is found by integrating a differential of area over appropriate limits. A vertical slice will have hight √x and width dx, so we have ...

dA = (√x)dx

A = ∫(√x)dx

The power rule can be used for the integration:

`\displaystyle A=\int_0^4{x^{(1)/(2)}}\,dx=\left[\frac{x^{(3)/(2)}}{(3)/(2)}\right]^4_0=(2)/(3)(4^(3)/(2))=(16)/(3)=\boxed{5(1)/(3)}`

The area of the region is 5 1/3 square units.

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Additional comment

You will notice that the area is bounded by a rectangle 4 units wide and 2 units high, for an area of 4·2 = 8. The area under the parabolic curve is 2/3 of that: 2/3·8 = 16/3 = 5 1/3 square units.

This fraction will be true for any area bounded by a parabola where the vertex is one of the corners of the rectangle.

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