To find the area of the region bounded by the parabola $y = 3x^2$, the tangent line at $(1, 3)$, and the x-axis, follow these steps:

Step 1: Find the equation of the tangent line.

The equation of a tangent line to a curve $y = f(x)$ at a point $(x_0, y_0)$ is given by:

$y - y_0 = f'(x_0)(x - x_0)$

Given the parabola $y = 3x^2$, its derivative is:

$f'(x) = \frac{d}{dx}(3x^2) = 6x$

At $x = 1$, the slope of the tangent is:

$f'(1) = 6(1) = 6$

So the tangent line at $(1, 3)$ is:

$y - 3 = 6(x - 1)$

Simplifying this, we get the equation of the tangent line:

$y = 6x - 3$

Step 2: Set up the integral to find the area.

The region is bounded by the parabola $y = 3x^2$, the tangent line $y = 6x - 3$, and the x-axis. We need to find the points of intersection between the parabola and the tangent line, and integrate between these points.

Set the equations equal to find the intersection points:

$3x^2 = 6x - 3$

Rearrange this into a quadratic equation:

$3x^2 - 6x + 3 = 0$

Divide by 3:

$x^2 - 2x + 1 = 0$

Factor the quadratic:

$(x - 1)^2 = 0$

Thus, $x = 1$. This means the parabola and the tangent intersect only at $x = 1$, the point of tangency.

Step 3: Find the limits of integration.

We are also interested in where the tangent line and parabola intersect the x-axis.

• For the tangent line $y = 6x - 3$, set $y = 0$ to find the x-intercept:

$0 = 6x - 3 \implies x = \frac{1}{2}$

So the tangent line intersects the x-axis at $x = \frac{1}{2}$.

• For the parabola $y = 3x^2$, set $y = 0$ to find the x-intercept:

$0 = 3x^2 \implies x = 0$

So the parabola intersects the x-axis at $x = 0$.

Step 4: Set up the integral for the area.

The area of the region between the parabola and the tangent line from $x = 0$ to $x = \frac{1}{2}$ is given by the difference between the equations of the tangent line and the parabola:

$\text{Area} = \int_0^{\frac{1}{2}} \left( (6x - 3) - 3x^2 \right) dx$

Step 5: Compute the integral.

Now, calculate the integral:

$\int_0^{\frac{1}{2}} \left( 6x - 3 - 3x^2 \right) dx$

This simplifies to:

$\int_0^{\frac{1}{2}} \left( 6x - 3 - 3x^2 \right) dx = \left[ 3x^2 - 3x - x^3 \right]_0^{\frac{1}{2}}$

Evaluate the definite integral:

At $x = \frac{1}{2}$:

$3\left( \frac{1}{2} \right)^2 - 3\left( \frac{1}{2} \right) - \left( \frac{1}{2} \right)^3 = 3\left( \frac{1}{4} \right) - 3\left( \frac{1}{2} \right) - \left( \frac{1}{8} \right) = \frac{3}{4} - \frac{3}{2} - \frac{1}{8}$

Simplify this expression:

$\frac{3}{4} - \frac{6}{4} - \frac{1}{8} = -\frac{3}{4} - \frac{1}{8} = -\frac{6}{8} - \frac{1}{8} = -\frac{7}{8}$

At $x = 0$, the integral is 0. Thus, the total area is:

$\text{Area} = -\left( -\frac{7}{8} \right) = \frac{7}{8}$

Final Answer:

The area of the region bounded by the parabola, the tangent line, and the x-axis is $\frac{7}{8}$ square units.

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Would you like any further details or explanations on the steps?

Here are 5 related questions:

1. How do you find the area between a curve and the x-axis in general?
2. What is the significance of the tangent line in optimization problems?
3. How do you derive the slope of a curve at a given point?
4. What is the geometric interpretation of a definite integral?
5. Can you calculate areas for bounded regions with more complex shapes?

Tip: When solving problems with curves and tangents, visualize the region to clearly understand the boundaries for integration.