Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$.
• Equation of a Tangent Line: The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.
• Parallel Lines: Two lines are parallel if and only if they have the same slope.

Step 1: Find the derivative of the given curve

We are given the curve $y = \int\limits_0^x {\left| t \right|dt,x \in R}$. We need to find $\frac{dy}{dx}$, which represents the slope of the tangent at any point $x$ on the curve. Applying the Fundamental Theorem of Calculus, Part 1, we have:

$\frac{dy}{dx} = \frac{d}{dx} \left( \int_0^x |t| dt \right) = |x|$

Thus, the slope of the tangent to the curve at any point $x$ is $|x|$.

Step 2: Determine the x-coordinates where the tangent is parallel to y = 2x

The given line is $y = 2x$. Its slope is $2$. Since the tangents to the curve are parallel to this line, their slopes must be equal. Therefore, we set the derivative equal to 2:

$|x| = 2$

Solving for $x$, we get two possible values:

$x = 2 \quad \text{or} \quad x = -2$

These are the x-coordinates of the points of tangency.

Step 3: Calculate the corresponding y-coordinates

Now we need to find the y-coordinates corresponding to $x = 2$ and $x = -2$ by plugging these values into the original equation $y = \int_0^x |t| dt$.

Case 1: x = 2

$y = \int_0^2 |t| dt$

Since $t$ ranges from $0$ to $2$, $t$ is non-negative, so $|t| = t$. Thus,

$y = \int_0^2 t dt = \left[ \frac{t^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2$

The first point of tangency is $(2, 2)$.

Case 2: x = -2

$y = \int_0^{-2} |t| dt$

We can reverse the limits of integration and negate the integral:

$y = -\int_{-2}^0 |t| dt$

Since $t$ ranges from $-2$ to $0$, $t$ is non-positive, so $|t| = -t$. Thus,

$y = -\int_{-2}^0 (-t) dt = \int_{-2}^0 t dt = \left[ \frac{t^2}{2} \right]_{-2}^0 = \frac{0^2}{2} - \frac{(-2)^2}{2} = 0 - \frac{4}{2} = -2$

The second point of tangency is $(-2, -2)$.

Step 4: Find the equations of the tangent lines

We have two points of tangency, $(2, 2)$ and $(-2, -2)$, and the slope $m = 2$. We use the point-slope form $y - y_1 = m(x - x_1)$.

Tangent 1: At (2, 2) with slope 2

$y - 2 = 2(x - 2)$ $y - 2 = 2x - 4$ $y = 2x - 2$

Tangent 2: At (-2, -2) with slope 2

$y - (-2) = 2(x - (-2))$ $y + 2 = 2(x + 2)$ $y + 2 = 2x + 4$ $y = 2x + 2$

Step 5: Determine the x-intercepts

To find the x-intercepts, we set $y = 0$ in the equations of the tangent lines and solve for $x$.

Tangent 1: y = 2x - 2

$0 = 2x - 2$ $2x = 2$ $x = 1$

Tangent 2: y = 2x + 2

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

The x-intercepts are $1$ and $-1$.

Common Mistakes & Tips

• Remember to consider both positive and negative cases when dealing with the absolute value function.
• When integrating with limits where the upper limit is less than the lower limit, reverse the limits and change the sign of the integral.
• Pay careful attention to signs when using the point-slope form.

Summary

We found the x-intercepts of the tangents to the curve $y = \int_0^x |t| dt$ that are parallel to the line $y = 2x$. We first found the derivative of the curve using the Fundamental Theorem of Calculus, then equated it to the slope of the given line to find the x-coordinates of the points of tangency. We then found the corresponding y-coordinates and used the point-slope form to find the equations of the tangent lines. Finally, we set $y = 0$ in the tangent equations to find the x-intercepts.

The final answer is $\boxed{\pm 1}$, which corresponds to option (A).