The intercepts on the x-axis made by tangents | vedclass

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(A) Given the curve $y = \int_{0}^{x} |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $\frac{dy}{dx} = |x|$.Since the tang

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$\pm 1$

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(A) Given the curve $y = \int_{0}^{x} |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $\frac{dy}{dx} = |x|$.Since the tangent is parallel to the line $y = 2x$,its slope must be $2$. Thus,$|x| = 2$,which gives $x = 2$ or $x = -2$.For $x = 2$,$y = \int_{0}^{2} |t| dt = \int_{0}^{2} t dt = [\frac{t^2}{2}]_{0}^{2} = 2$. The point is $(2, 2)$.The equation of the tangent at $(2, 2)$ is $y - 2 = 2(x - 2) \Rightarrow y = 2x - 2$. The $x$-intercept is found by setting $y = 0$,giving $2x = 2$,so $x = 1$.For $x = -2$,$y = \int_{0}^{-2} |t| dt = \int_{0}^{-2} (-t) dt = [-\frac{t^2}{2}]_{0}^{-2} = -2$. The point is $(-2, -2)$.The equation of the tangent at $(-2, -2)$ is $y - (-2) = 2(x - (-2)) \Rightarrow y + 2 = 2x + 4 \Rightarrow y = 2x + 2$. The $x$-intercept is found by setting $y = 0$,giving $2x = -2$,so $x = -1$.Thus,the $x$-intercepts are $\pm 1$.

The intercepts on the $x$-axis made by tangents to the curve $y = \int_{0}^{x} |t| dt, x \in R$ which are parallel to the line $y = 2x$ are equal to:

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