# Find the area of the region bounded by the parabola $y = 2x^2$, the tangent line to this parabola at $(3, 18)$, and the $x$-axis.

$$f(x) = 2x^2 \gets$$ this is the parabola

$$f(3) = 2 \times 9 = 18 \to$$ the parabola passes through $$A (3 ; 18$$), so its tangent line does too.

$$f'(x) = 4x \gets$$ this is the derivative

…and the derivative is the slope of the tangent line to the curve at $$x$$

$$f'(3) = 4 \times 3 = 12 \gets$$ this is the slope of the tangent line to the curve at $$x = 3$$

Equation of the tangent line

The typical equation of a line is: $$y = mx + b$$, where $$m$$ is the slope and $$b$$ is the $$y$$-intercept.

We know that the slope of the tangent line is $$12$$.

The equation of the tangent line becomes $$y = 12x + b$$.

The tangent line passes through $$A (3 ; 18)$$, so these coordinates must satisfy the equation of the tangent line.

$$y = 12x + b$$

$$b = y - 12x \to$$ I substitute $$x$$ and $$y$$ by the coordinates of the point $$A (3 ; 18)$$.

$$b = 18 - 36 = - 18$$

$$\to$$ The equation of the tangent line is $$y = 12x - 18$$.

Intersection between the tangent line to the curve and the $$x$$-axis: $$\to$$ when $$y = 0$$. \begin{align} y &= 12x - 18 \to \text{when } y = 0 \\ 12x - 18 &= 0 \\ 12x &= 18 \\ x &= 3/2 \end{align}

$$\to$$ Point $$B (3/2 ; 0)$$

Intersection between the vertical line passes through the point $$A$$ and the $$x$$-axis: $$\to$$ when $$x = 3$$.

$$\to$$ Point $$C (3 ; 0)$$

The equation of the vertical line is $$x = 3$$.

Area of the region bounded by the parabola $$y = 2x^2$$, the tangent line to this parabola at $$(3; 18)$$, and the $$x$$-axis.

$$=$$ (area of the region bounded by the parabola $$y = 2x^2$$ and the $$x$$-axis) minus (area of the triangle $$ABC$$)

$$=$$ [integral (from $$0$$ to $$3$$) of the parabola] minus $$[(x_C-x_B)\cdot(y_A-y_C)/2]$$

\begin{align} &= \int_0^3 2x^2 dx -\frac{(x_C-x_B)(y_A-y_C}{2} \\ &= \left. \frac23 x^3 \right|_0^3 -\frac{(3-3/2)(18-0)}{2} \\ &= \frac23 \cdot 3^2 -(6/2 - 3/2)\cdot9 \\ &= \frac23 \cdot 27 -\frac32 \cdot 9 \\ &= 18 - \frac{27}{2} \\ &= \frac{36}{2} -\frac{27}{2} \\ &= \frac92 \text{ square units} \end{align}

• Well explained. The method, answer are correct. – André Nicolas Oct 31 '15 at 0:33
• Hint: to check your own work on such integrals, make a sketch of the situation. – B. Pasternak Oct 31 '15 at 8:27

$$y=2x^2$$ has derivative $$y'=4x$$. So the derivative will be positive for all $$x>0$$. This means the tangent line will intersect the x-axis at some point $$x_a for a given x.

This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $$x which is made up of only area under the parabola, and a region $$x>x_a$$ where area is from the area between the parabola and the tangent to $$(3,18)$$.

The equation of a tangent line at $$(x_0,y_0)$$ is $$\frac{y-f(x_0)}{x-x_0}=f'(x_0)$$

Or : $$y=f'(x_0)(x-x_0)+f(x_0)$$

The x intercept happens where $$y=0$$.

Requiring $$y=0$$ implies an x intercept of $$x_c=\frac{-f(x_0)}{f'(x_0)}+x_0$$

So from the above arguments with $$x_0=3$$:

$$A=\int_0^{x_c}2x^2dx+\int_{x_c}^32x^2-(12x-18)dx$$

But this can be simplified. The area under the tangent line is a triangle. So the the parabola can be integrated ignoring the tangent line, and then subtracting the area of the triangle.

From the above, we know $$(x_0-x_c)=\frac{f(x_0)}{f'(x_0)}$$, the base of the triangle. The height is just $$f(x_0)$$.

So the area of the triangle is $$A_t=\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$$

So:

$$A=\int_0^{x_0}f(x)dx-\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$$

and solve for $$x_0=3$$.

In this form, an expression can be found for $$x_0$$ which extremizes the area.