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

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(A) Given the curve $y = f(x) = \int_0^x |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $f'(x) = |x|$. The tangent is par

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$\pm 1$

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(A) Given the curve $y = f(x) = \int_0^x |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $f'(x) = |x|$. The tangent is parallel to the line $y = 2x$,so its slope is $2$. Thus,$|x| = 2$,which implies $x = 2$ or $x = -2$. Case $1$: If $x = 2$,then $y = \int_0^2 t dt = [\frac{t^2}{2}]_0^2 = 2$. The point is $(2, 2)$. The equation of the tangent is $y - 2 = 2(x - 2) \implies y = 2x - 2$. The $x$-intercept is found by setting $y = 0$,giving $0 = 2x - 2 \implies x = 1$. Case $2$: If $x = -2$,then $y = \int_0^{-2} |t| dt = \int_0^{-2} (-t) dt = [-\frac{t^2}{2}]_0^{-2} = -2$. The point is $(-2, -2)$. The equation of the tangent is $y - (-2) = 2(x - (-2)) \implies y + 2 = 2x + 4 \implies y = 2x + 2$. The $x$-intercept is found by setting $y = 0$,giving $0 = 2x + 2 \implies x = -1$. Therefore,the $x$-intercepts are $\pm 1$.

The intercepts on the $x$-axis made by the 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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