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

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(B) Given the curve $y(x) = \int_0^x (2t^2 - 15t + 10) dt$. By the Fundamental Theorem of Calculus,the derivative is $y'(x) = 2x^2 - 15x + 10$.At poin

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$406$

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(B) Given the curve $y(x) = \int_0^x (2t^2 - 15t + 10) dt$. By the Fundamental Theorem of Calculus,the derivative is $y'(x) = 2x^2 - 15x + 10$.At point $(a, b)$,the slope of the tangent is $m_T = y'(a) = 2a^2 - 15a + 10$.The slope of the normal is $m_N = -\frac{1}{y'(a)} = -\frac{1}{2a^2 - 15a + 10}$.The given line is $x + 3y = -5$,which can be written as $y = -\frac{1}{3}x - \frac{5}{3}$. The slope of this line is $m_L = -\frac{1}{3}$.Since the normal is parallel to the line,$m_N = m_L$,so $-\frac{1}{2a^2 - 15a + 10} = -\frac{1}{3}$.This implies $2a^2 - 15a + 10 = 3$,so $2a^2 - 15a + 7 = 0$.Factoring the quadratic: $(2a - 1)(a - 7) = 0$. Since $a > 1$,we have $a = 7$.Now,$b = y(a) = \int_0^7 (2t^2 - 15t + 10) dt = [\frac{2}{3}t^3 - \frac{15}{2}t^2 + 10t]_0^7$.$b = \frac{2}{3}(343) - \frac{15}{2}(49) + 10(7) = \frac{686}{3} - \frac{735}{2} + 70 = \frac{1372 - 2205 + 420}{6} = -\frac{413}{6}$.We need $|a + 6b| = |7 + 6(-\frac{413}{6})| = |7 - 413| = |-406| = 406$.

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 $x + 3y = -5$,where $a > 1$,then the value of $|a + 6b|$ is equal to...........

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