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Is my answer correct?

$f(x) = 2x^2 \gets$ 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}

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    $\begingroup$ Well explained. The method, answer are correct. $\endgroup$ Oct 31, 2015 at 0:33
  • $\begingroup$ Hint: to check your own work on such integrals, make a sketch of the situation. $\endgroup$ Oct 31, 2015 at 8:27

1 Answer 1

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$y=2x^2$ 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 $x_a<x$ 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 $x<x_a$ 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.

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