If the normal to the curve y(x) = int_0^x (2t | vedclass

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(B) Given $y(x) = \int_0^x (2t^2 - 15t + 10) dt$. The slope of the tangent at $(a, b)$ is $y'(a) = 2a^2 - 15a + 10$. The slope of the line $x + 3y = -

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$406$

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(B) Given $y(x) = \int_0^x (2t^2 - 15t + 10) dt$. The slope of the tangent at $(a, b)$ is $y'(a) = 2a^2 - 15a + 10$. The slope of the line $x + 3y = -5$ is $m = -1/3$. Since the normal is parallel to the line,the slope of the normal is $-1/3$. The slope of the tangent is the negative reciprocal of the slope of the normal,so $y'(a) = 3$. $2a^2 - 15a + 10 = 3 \Rightarrow 2a^2 - 15a + 7 = 0$. $(2a - 1)(a - 7) = 0$. Since $a > 1$,we have $a = 7$. Now,$b = y(7) = \int_0^7 (2t^2 - 15t + 10) dt = [\frac{2t^3}{3} - \frac{15t^2}{2} + 10t]_0^7$. $b = \frac{2(343)}{3} - \frac{15(49)}{2} + 70 = \frac{686}{3} - \frac{735}{2} + 70 = \frac{1372 - 2205 + 420}{6} = -\frac{413}{6}$. Thus,$6b = -413$. $|a + 6b| = |7 - 413| = |-406| = 406$.

If the normal to the curve $y(x) = \int_0^x (2t^2 - 15t + 10) dt$ at a point $(a, b)$ is parallel to the line $x + 3y = -5$ for $a > 1$,then the value of $|a + 6b|$ is equal to ...........

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