Question

find the area of the region bounded by the parabola y = 3x^2, the tangent line to this parabola at (3, 27), and the x-axis.

Answer 1

The area of the region bounded by the parabola y = 3x^2, the tangent line at (3, 27), and the x-axis is 13.5 square units.

Step-by-step explanation:

The area of the region bounded by the parabola y = 3x^2, the tangent line to this parabola at (3, 27), and the x-axis can be found by calculating the area of the triangle formed by the tangent line and the x-axis and subtracting it from the area under the curve.

To find the area of the triangle, we need to find the base and height of the triangle. The base is the difference between the x-coordinate of the point where the tangent line intersects the x-axis (which is 3) and the x-coordinate of the vertex of the parabola (which is 0). The height is the y-coordinate of the point where the tangent line intersects the parabola (which is 27). Using these values, we can calculate the area of the triangle as: Area = (1/2) * base * height = (1/2) * (3 - 0) * 27 = 40.5 square units.

To find the area under the curve, we need to integrate the function y = 3x^2 from x = 0 to x = 3. The integral of y = 3x^2 is given by the formula: Area = integral(3x^2)dx = x^3 from x = 0 to x = 3 = 3^3 - 0^3 = 27 square units.

Finally, we can find the area of the region bounded by the parabola, the tangent line, and the x-axis by subtracting the area of the triangle from the area under the curve: Area = 27 - 40.5 = -13.5 square units. However, since the area cannot be negative, we take the absolute value of the result, giving us the final answer: Area = 13.5 square units.

Answer 2

The area of the region bounded by the parabola
`\(y = 3x^2\)`, the tangent line at
`\((3, 27)\)`, and the x-axis is
`\(27\)` square units.

To find the area of the region bounded by the parabola
`\(y = 3x^2\)`, the tangent line at
`\((3, 27)\)`, and the x-axis, you'll need to find the x-coordinate where the tangent line intersects the parabola. Then, you can set up the integral to find the area between these curves.

First, let's find the equation of the tangent line to the parabola
`\(y = 3x^2\)` at the point
`\((3, 27)\)`. The derivative of
`\(y = 3x^2\)` is
`\(y' = 6x\)`, which gives the slope of the tangent line at any point
`\(x\)`.

At
`\(x = 3\)`, the slope of the tangent line is
`\(6 * 3 = 18\)`. The equation of the tangent line in point-slope form is:

`\(y - y_1 = m(x - x_1)\),`

where
`\(m\)` is the slope and
`\((x_1, y_1)\)` is the point on the line.

Using
`\((x_1, y_1) = (3, 27)\)` and
`\(m = 18\)`:

`\[y - 27 = 18(x - 3)\]`

`\[y - 27 = 18x - 54\]`

`\[y = 18x - 27\]`

To find the x-coordinate where the tangent line intersects the parabola, set the equations of the parabola and the tangent line equal to each other:

`\[3x^2 = 18x - 27\]`

`\[3x^2 - 18x + 27 = 0\]`

`\[x^2 - 6x + 9 = 0\]`

`\((x - 3)^2 = 0\)`

This gives us a repeated root of
`\(x = 3\)`, meaning the tangent line intersects the parabola at
`\(x = 3\)`.

Now, to find the area bounded by these curves, integrate the difference between the curves from
`\(x = 0\)` to
`\(x = 3\)`. The curves are
`\(y = 3x^2\)` (the parabola) and
`\(y = 0\)` (the x-axis).

The area
`\(A\)` is given by:

`\[A = \int_0^3 (3x^2 - 0) \, dx\]`

`\[A = \int_0^3 3x^2 \, dx\]`

`\[A = \left[x^3\right]_0^3\]`

`\[A = 3^3 - 0^3\]`

`\[A = 27\]`

Therefore, the answer is
`\(27\)` square units.