Curves

To find the slope of the normal to the curve at , we follow these steps:

1. Determine the derivative of :
The derivative gives the slope of the tangent line at any point.

2. Find the slope of the tangent at :
Substitute into the derivative.

3. Calculate the slope of the normal:
The slope of the normal line is the negative reciprocal of the slope of the tangent line.

If the slope of the tangent is , then

Thus, the slope of the normal to the curve at is .

The problem requires finding the area bounded by the curve , the x-axis (y = 0), x = 0, and x = 5.

First, analyze the curve:

The critical point is at . The expression forms a V-shaped graph.

Calculate the area between x = 0 and x = 5 by dividing it into two regions using x = 2.

Summing both areas gives:

Final Area = square units.

Hence, the area bounded by the curve is 6.5 square units.

To find the area enclosed between the curve and the line , we begin by identifying the points of intersection. Solving both equations simultaneously:

and

Substitute into :

This gives or . Hence, or , giving the points (0,0) and (4,4).

Now, calculate the area between the curve and line using integration:

1. The parabola:

2. Line:

Determine the area from to :

Evaluate each integral separately:

Hence, the area is square units.

To find the area bounded by the curves and , we first need to determine their points of intersection. We set the equations equal:

This simplifies to:

Factor out :

Further factorize:

Thus, the points of intersection are , , and .

We will consider the area between the curves from to , because the region is symmetric about the origin.

The area is given by the integral of the difference of the functions:

Calculate the integral:

Compute :

evaluated from 0 to 1 is

Compute :

evaluated from 0 to 1 is

Thus, the area is:

Therefore, the area of the region bounded by the curves is square units.